Algebra 2 · F.BF.B.5 · F.LE.A.4

Solving by taking a logarithm of both sides

The unknown sits up in the exponent, and the two sides are not powers of one base. How do you bring it down to where it can be solved for?

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Laws of logarithmsAlgebra 2

Where you are 11 in Logarithms

Functions13

Logarithms11

▴ 2 above Expanding and condensing logarithms Laws of logarithms

Solving by taking a logarithm of both sides

Solving exponential equations by matching bases Evaluating logarithms, common and natural ▾ Show all 11 (4 more)

When you reach for it

Some equations with the unknown in the exponent give way to a trick: write both sides as powers of one base and set the exponents equal, which is solving by matching bases. That trick needs luck. Look at

7^x = 40

There is no whole-number power of 7 that makes 40, since 7^1 = 7 and 7^2 = 49, and 40 is not a power of 7 in any tidy way. The answer is some number between 1 and 2, nearer 2, and trying decimals by hand would only ever close in on it.

The way out is the power law of logarithms, \log_b\left(M^p\right) = p\log_b M. It takes an exponent and turns it into a factor. So if you take the logarithm of both sides of an equation, an unknown that was up in the exponent comes down to the main line, where ordinary algebra can reach it.

Reach for it whenever the unknown is in an exponent and the bases do not match: to find when a balance reaches a target, when a population passes a size, when a warm loaf cools to a temperature, or how long a radioactive sample takes to lose half its mass. In the lessons on Compound interest and compounding periods and Doubling time and half life questions like these were answered by trying whole numbers of periods one after another. With logarithms they are answered exactly.

The procedure

Why you may take a logarithm of both sides

If two numbers are equal, then doing the same thing to both gives equal results. That is all an equation step ever is. A logarithm is a function: each positive input has exactly one output. So if M = N and both are positive, then

\log M = \log N

The catch is the word positive. A logarithm only accepts positive inputs, so each side must be positive before the logarithm is taken. A power b^x with a positive base is always positive, which is why an equation of the form b^x = c with c > 0 is always safe to log.

The step also runs backwards. The function \log is one-to-one (two different positive inputs never share an output, because its graph is always rising), so \log M = \log N forces M = N. Nothing is gained or lost in taking the logarithm: the new equation has exactly the same solutions as the old one.

Bringing the exponent down

Start from 7^x = 40. Both sides are positive, so take the common logarithm of each:

\log\left(7^x\right) = \log 40

The power law turns the left side into x \log 7:

x \log 7 = \log 40

Now x is multiplied by a number, \log 7 \approx 0.8451, and dividing by it is allowed because it is not zero (7 \neq 1, and only \log 1 is 0):

x = \frac{\log 40}{\log 7} \approx \frac{1.6021}{0.8451} \approx 1.896

The exact answer is the fraction \displaystyle \frac{\log 40}{\log 7}. The decimal is an approximation, and it should be made only at the very end, by entering the whole fraction into a calculator in one go; rounding \log 40 and \log 7 first and then dividing can shift the last digit. It agrees with the estimate: 1.896 lies between 1 and 2 and is nearer 2, as 40 is nearer 49 than 7.

Figure 1The curve y = 7^x meets the line y = 40 exactly once. The logarithm finds where, \displaystyle x = \frac{\log 40}{\log 7} \approx 1.896.

Which logarithm to take

Any base works, because the power law holds in every base. Calculators carry two, the common logarithm \log (base 10) and the natural logarithm \ln (base e), so in practice it is one of those. They give the same answer:

x = \frac{\ln 40}{\ln 7} \approx \frac{3.6889}{1.9459} \approx 1.896

Use \log when the power has base 10, since \log\left(10^x\right) = x exactly; use \ln when the power has base e, since \ln\left(e^x\right) = x; for any other base, either one. The answer also has a third name: 7^x = 40 says x = \log_7 40 in logarithmic form, and the work above is the reason that \log_7 40 can be found from the two logarithms on a calculator, which is the subject of Change of base.

Isolate the power first

The logarithm of a sum is not the sum of the logarithms, so there is no law that opens up \log\left(5 \cdot 3^x + 8\right). Before taking logarithms, use ordinary inverse operations to get the power by itself on one side: subtract what is added, then divide by what multiplies. A factor in front may also be left in place, since the product law does split a product, \log\left(5 \cdot 3^x\right) = \log 5 + x\log 3, but anything added or subtracted must go first.

When the unknown is on both sides

If the unknown appears in two exponents with different bases, as in 3^{x + 2} = 5^x, taking logarithms brings both exponents down at once. What is left is a linear equation in x, with logarithms as its coefficients: gather the terms in x on one side, factor out x, and divide.

Worked examples

The lesson is taught in two parts. Examples 1 to 4 belong to the first: one power, isolated and then logged, with the exponent a single letter or a linear expression. Examples 5 to 9 belong to the second: the unknown in two exponents, and the growth and decay models met earlier, now solved for time.

Example 1

One power on its own

Solve each equation. Give the exact answer and the value to three decimal places.

(a) 6^x = 75 (b) 10^x = 0.04 (c) 0.8^x = 0.3

Solution

(a) Both sides are positive. Taking the common logarithm and using the power law,

\begin{aligned} x \log 6 &= \log 75 \\ x &= \frac{\log 75}{\log 6} \approx 2.410 \end{aligned}

Check by size: 6^2 = 36 and 6^3 = 216, and 75 lies between them, closer to 36, so x should be between 2 and 3 and nearer 2. It is.

(b) The base is 10, so the common logarithm undoes the power directly: \log\left(10^x\right) = x. Hence

x = \log 0.04 \approx -1.398

The answer is negative, as it must be: 0.04 is less than 1 = 10^0, and a power of 10 is below 1 exactly when its exponent is below 0.

\begin{aligned} x \log 0.8 &= \log 0.3 \\ x &= \frac{\log 0.3}{\log 0.8} \approx \frac{-0.5229}{-0.0969} \approx 5.396 \end{aligned}

Both logarithms are negative, because 0.3 and 0.8 are both less than 1, and the quotient of two negatives is positive. That fits: 0.8^x shrinks as x grows, so it takes several factors of 0.8 to fall from 1 to 0.3.

Example 2

A number in front, two ways

A town has 3{,}000 people, and its population grows by 6\% a year. Its population in thousands after t years is modeled by P = 3(1.06)^t. When does the population reach 10{,}000?

Solution

Ten thousand people is P = 10, so the equation is

3(1.06)^t = 10

First way: divide, then log. Dividing both sides by 3 isolates the power, \displaystyle (1.06)^t = \frac{10}{3}. Both sides are positive, so take logarithms:

\begin{aligned} t \log 1.06 &= \log \tfrac{10}{3} \\ t &= \frac{\log \frac{10}{3}}{\log 1.06} \approx 20.66 \end{aligned}

Second way: log the product. The left side is a product, and the product law splits it: \log\left(3(1.06)^t\right) = \log 3 + t\log 1.06. Since \log 10 = 1,

\begin{aligned} \log 3 + t\log 1.06 &= 1 \\ t &= \frac{1 - \log 3}{\log 1.06} \approx 20.66 \end{aligned}

The two exact answers are the same number, because by the quotient law \displaystyle \log \frac{10}{3} = \log 10 - \log 3 = 1 - \log 3.

So the population reaches 10{,}000 about 20.66 years from now, that is, during the 21st year. As a check, 3(1.06)^{20} \approx 9.62 and 3(1.06)^{21} \approx 10.20: just under 10 thousand after 20 years and just over it after 21.

Example 3

An exponent that is an expression, and base e

Solve, giving exact answers and three decimal places.

(a) 2^{3x - 1} = 19 (b) 6e^{0.4x} = 45

Solution

(a) The power is already alone. The power law brings down the whole exponent 3x - 1, so it must be kept together in brackets:

\begin{aligned} (3x - 1)\ln 2 &= \ln 19 \\ 3x - 1 &= \frac{\ln 19}{\ln 2} \\ x &= \frac{1}{3}\left(1 + \frac{\ln 19}{\ln 2}\right) \approx 1.749 \end{aligned}

Dividing by \ln 2 before expanding keeps the arithmetic short: the exponent 3x - 1 is about 4.248, a little more than 4, as 19 is a little more than 2^4 = 16.

(b) Divide by 6 to isolate the power: e^{0.4x} = 7.5. The base is e, so take the natural logarithm, which undoes it exactly, \ln\left(e^{0.4x}\right) = 0.4x:

\begin{aligned} 0.4x &= \ln 7.5 \\ x &= \frac{\ln 7.5}{0.4} \approx 5.037 \end{aligned}

Using \log here would also work, giving 0.4x \log e = \log 7.5, but it drags in the extra number \log e. The natural logarithm is the one that matches base e.

Example 4

When there is no solution

Solve (a) 3^x + 8 = 5 and (b) -2 \cdot 5^x = -14.

Solution

(a) Subtracting 8 isolates the power: 3^x = -3. The right side is negative, and 3^x is positive for every real x, so no x works and the equation has no solution. Taking logarithms would have been the wrong move anyway, since \log(-3) does not exist. The graph shows the same thing: every output of 3^x + 8 is above 8, so the curve never comes down to the line y = 5.

Figure 2Every output of 3^x + 8 is above 8, so the equation 3^x + 8 = 5 has no solution.

(b) Here the negative signs cancel. Dividing both sides by -2 gives 5^x = 7, and now both sides are positive:

x = \frac{\ln 7}{\ln 5} \approx 1.209

The lesson of the pair is that positivity is judged after the power is isolated, not by looking at the signs in the original equation.

Example 5

Different bases on the two sides

Solve 3^{x + 2} = 5^x.

Solution

No base can be matched here, since 3 and 5 are not powers of a common number. Both sides are positive for every x, so take natural logarithms and bring both exponents down:

(x + 2)\ln 3 = x \ln 5

This is a linear equation in x whose coefficients happen to be logarithms. Expand by the distributive property, then gather the terms in x on one side:

\begin{aligned} x\ln 3 + 2\ln 3 &= x\ln 5 \\ 2\ln 3 &= x\ln 5 - x\ln 3 \\ 2\ln 3 &= x(\ln 5 - \ln 3) \end{aligned}

The bracket \ln 5 - \ln 3 is positive, since 5 > 3, so dividing by it is allowed:

x = \frac{2\ln 3}{\ln 5 - \ln 3} = \frac{\ln 9}{\ln \frac{5}{3}} \approx 4.301

The second form uses the power law on top (2\ln 3 = \ln 9) and the quotient law below. It also reads cleanly: dividing both sides of 3^{x+2} = 5^x by 3^x gives \displaystyle 9 = \left(\frac{5}{3}\right)^x. Check: with x \approx 4.301 both sides are about e^{6.92}, roughly 1{,}000.

Example 6

Savings solved for time

4{,}500 dollars is deposited at 3.8\% a year, compounded monthly. After how many months does the balance first reach 6{,}000 dollars?

Solution

The rate per month is \displaystyle \frac{0.038}{12}, so after t years, which is 12t months, the balance in dollars is

4500\left(1 + \frac{0.038}{12}\right)^{12t}

Set it equal to 6000 and divide by 4500 to isolate the power:

\left(1 + \frac{0.038}{12}\right)^{12t} = \frac{4}{3}

Take natural logarithms and bring the exponent 12t down:

\begin{aligned} 12t \ln\left(1 + \tfrac{0.038}{12}\right) &= \ln \tfrac{4}{3} \\ 12t &= \frac{\ln \frac{4}{3}}{\ln\left(1 + \frac{0.038}{12}\right)} \approx 90.99 \end{aligned}

So the equation is satisfied after about 90.99 months, t \approx 7.58 years. But interest is only added at the end of each month, so the balance moves in steps and is never exactly 6{,}000 dollars. After 90 months it is about 5{,}981.24 dollars, still short, and after 91 months about 6{,}000.18 dollars. The balance first reaches 6{,}000 dollars after 91 months, which is 7 years and 7 months.

Example 7

A loaf cooling on a rack

A loaf of bread comes out of the oven at 95 degrees Celsius and is set on a rack in a kitchen at 25 degrees. Its temperature after t minutes is T = 25 + 70e^{-0.04t} degrees. It slices cleanly once it is down to 40 degrees. When is that?

Solution

At t = 0 the model gives 25 + 70 = 95, as it should. Set T = 40 and isolate the power, subtracting 25 and then dividing by 70:

\begin{aligned} 70e^{-0.04t} &= 15 \\ e^{-0.04t} &= \frac{3}{14} \end{aligned}

Both sides are positive. The base is e, so take the natural logarithm:

\begin{aligned} -0.04t &= \ln \tfrac{3}{14} \\ t &= \frac{\ln \frac{14}{3}}{0.04} \approx 38.51 \end{aligned}

The second line divides by -0.04 and uses \displaystyle -\ln \frac{3}{14} = \ln \frac{14}{3}, the reciprocal rule. The loaf reaches 40 degrees after about 38.51 minutes, a little over half an hour.

Example 8

Finding the rate, then the half-life

A radioactive sample has a mass of 80 grams, and 10 days later its mass is 65 grams. Its mass after t days is modeled by A = 80e^{kt}. Find k, and find the half-life of the substance.

Solution

Here the unknown in the exponent is first the constant k. Put t = 10 and A = 65:

\begin{aligned} 80e^{10k} &= 65 \\ e^{10k} &= \frac{13}{16} \\ 10k &= \ln \tfrac{13}{16} \\ k &= \frac{\ln \frac{13}{16}}{10} \approx -0.02076 \end{aligned}

The value of k is negative, as decay requires. The half-life is the time T at which half the mass is left, 80e^{kT} = 40. Dividing by 80 gives \displaystyle e^{kT} = \frac{1}{2}, and taking natural logarithms,

kT = \ln \tfrac{1}{2} = -\ln 2

Dividing by k, and using the exact value of k,

T = \frac{-\ln 2}{k} = \frac{10\ln 2}{\ln \frac{16}{13}} \approx 33.38

The half-life is about 33.38 days. It is sensible: 65 grams is still well over half of 80 after 10 days, so halving should take several times 10 days. Notice that the starting mass 80 dropped out of the half-life; any amount of this substance halves in the same time.

Example 9

A hidden quadratic

Solve 25^x - 2 \cdot 5^x - 8 = 0.

Solution

The terms cannot be logged one at a time, since logarithms do not split a sum. But 25^x = \left(5^2\right)^x = 5^{2x} = \left(5^x\right)^2, by the power of a power rule and the commutative property of multiplication (2x = x \cdot 2). So with u = 5^x the equation is a quadratic in u:

u^2 - 2u - 8 = 0

It factors as (u - 4)(u + 2) = 0, so by the Solving quadratics by factoring u = 4 or u = -2. Back in terms of x, that is 5^x = 4 or 5^x = -2. The second has no solution, because 5^x is always positive. The first does:

x = \frac{\ln 4}{\ln 5} \approx 0.861

Check: 5^x = 4 gives 25^x = 16, and 16 - 2 \cdot 4 - 8 = 0.

Boundaries

Each side must be positive when the logarithm is taken. A logarithm accepts only positive inputs. Since b^x > 0 for every real x when b > 0, an equation that isolates to b^x = c with c \leq 0 has no solution at all, as in Example 4. The test is made after the power is isolated.

The base must be positive and not 1. The final division is by \log b, which is 0 when b = 1; and indeed 1^x = c is true for every x when c = 1 and for none otherwise, so there is nothing to solve for. A negative base such as (-2)^x is not defined for every real x (for instance (-2)^{1/2} is not a real number), so the method, and the power law it rests on, do not apply.

Only a single power can be logged. The logarithm of a sum has no law. Constants added to the power must be moved first, and an equation with two different powers added together, such as 25^x - 2 \cdot 5^x - 8 = 0 in Example 9, needs a substitution before logarithms can help.

The unknown must be only in exponents. In an equation like 2^x = 3x, taking logarithms gives x\ln 2 = \ln 3 + \ln x, which still has x in two different forms. No algebra with logarithms isolates x there; such equations are solved by graphing or by numerical approximation.

A model’s answer may need rounding to the model’s steps. When a quantity changes only at the end of each period, as interest compounded monthly does, the exact solution of the equation falls between periods, and the answer to the question is the next whole period (Example 6).

Where this goes next

The quotient \displaystyle \frac{\log c}{\log b} that solves b^x = c is also the value of \log_b c, and turning that observation into a rule is Change of base. The reverse move, raising a base to both sides to undo a logarithm, solves equations such as \log_3(x + 1) = 2, where answers must be tested against the original logarithms; that is Solving logarithmic equations.

Taking a logarithm of the output of an exponential model turns its curve into a straight line, because \log\left(ab^t\right) = \log a + t\log b is linear in t. Plotting data that way, and reading a growth factor from the slope, is the idea of Semi-log plots: straightening exponential data with logarithms. Throughout the rest of the study of growth and decay, in compound interest, continuous growth and half-lives, every question of the form “how long until” is answered by the move in this lesson.

Exercises

Part 1: one power on its own

Solve. Give the exact answer and the value to three decimal places.

4^x = 30
10^x = 520
e^x = 0.25
0.6^x = 0.05

Part 1: isolate the power first

Solve. Give the exact answer and the value to three decimal places, or show there is no solution.

5 \cdot 3^x + 8 = 70
5e^x - 2 = 18
2^{x + 5} = 9
8e^{-0.3x} = 2
7^{2x - 1} = 60
6 + 2^x = 4
17 - 6 \cdot 2^x = 3

Part 2: the unknown in two exponents

Solve. Give the exact answer and the value to three decimal places.

2^{x + 4} = 3^x
6^{x - 1} = 2^x
e^{2x} = 3^{x + 1}
e^{2x} - 5e^x + 6 = 0

Part 2: models solved for time

A city of 85{,}000 people grows by 2.1\% a year, so its population in thousands after t years is 85(1.021)^t. When does it reach 100{,}000 people?
2{,}500 dollars is deposited at 4.5\% a year, compounded quarterly. Write the balance after t years, find when the equation says the balance doubles, and find after how many quarters it first reaches 5{,}000 dollars.
1{,}200 dollars is deposited at 5\% a year, compounded continuously, so the balance is 1200e^{0.05t} dollars after t years. When does it reach 2{,}000 dollars?
A delivery van costs 21{,}000 dollars new and loses 12\% of its value each year, so its value after t years is 21000(0.88)^t dollars. When is it worth 7{,}000 dollars?
Town A has 40{,}000 people and grows by 3\% a year. Town B has 65{,}000 people and grows by 1\% a year. When will the two towns have the same population?
A sample of a radioactive substance has a mass of 60 grams, and 6 hours later its mass is 50.4 grams. Using the model A = 60e^{kt} with t in hours, find k and the half-life.

Reasoning

Show, with one pair of numbers, that \log(a + b) is not in general equal to \log a + \log b. Then explain why, in solving 5 \cdot 3^x + 8 = 70, the 8 must be moved before logarithms are taken, while the factor 5 may stay.
Prove that for positive c and a positive base b \neq 1, \displaystyle \frac{\log c}{\log b} = \frac{\ln c}{\ln b}, so it never matters which of the two logarithms is used to solve b^x = c.
Explain why, for a base b > 0 with b \neq 1, the equation b^x = c has exactly one solution when c > 0 and none when c \leq 0.
Is \displaystyle \frac{\ln 40}{\ln 7} the same number as \displaystyle \ln \frac{40}{7}? Work out both to three decimal places, and say which one solves 7^x = 40.

Answers

\displaystyle x = \frac{\ln 30}{\ln 4} \approx 2.453. Taking logarithms, x\ln 4 = \ln 30. Check: 4^2 = 16 and 4^3 = 64.
x = \log 520 \approx 2.716. The common logarithm undoes base 10 exactly.
x = \ln 0.25 \approx -1.386. The natural logarithm undoes base e; negative because 0.25 < 1.
\displaystyle x = \frac{\ln 0.05}{\ln 0.6} \approx 5.864. Both logarithms are negative, so the quotient is positive.
\displaystyle x = \frac{\ln 12.4}{\ln 3} \approx 2.292. Subtract 8 and divide by 5: 3^x = 12.4.
x = \ln 4 \approx 1.386. Add 2 and divide by 5: e^x = 4.
\displaystyle x = \frac{\ln 9}{\ln 2} - 5 \approx -1.830. Taking logarithms, (x + 5)\ln 2 = \ln 9, so x + 5 \approx 3.170.
\displaystyle x = \frac{\ln 4}{0.3} \approx 4.621. Divide by 8: \displaystyle e^{-0.3x} = \frac{1}{4}, so \displaystyle -0.3x = \ln \frac{1}{4} = -\ln 4.
\displaystyle x = \frac{1}{2}\left(1 + \frac{\ln 60}{\ln 7}\right) \approx 1.552. Taking logarithms, (2x - 1)\ln 7 = \ln 60, so 2x - 1 \approx 2.104.
No solution. Subtracting 6 gives 2^x = -2, and 2^x is positive for every x.
\displaystyle x = \frac{\ln \frac{7}{3}}{\ln 2} \approx 1.222. Subtract 17: -6 \cdot 2^x = -14, then divide by -6: \displaystyle 2^x = \frac{7}{3}.
\displaystyle x = \frac{4\ln 2}{\ln 3 - \ln 2} \approx 6.838. From (x + 4)\ln 2 = x\ln 3, gather: 4\ln 2 = x(\ln 3 - \ln 2).
\displaystyle x = \frac{\ln 6}{\ln 3} \approx 1.631. From (x - 1)\ln 6 = x\ln 2: x(\ln 6 - \ln 2) = \ln 6, and \ln 6 - \ln 2 = \ln 3 by the quotient law.
\displaystyle x = \frac{\ln 3}{2 - \ln 3} \approx 1.219. Natural logarithms give 2x = (x + 1)\ln 3, so x(2 - \ln 3) = \ln 3; the bracket is positive since \ln 3 < 2.
x = \ln 2 \approx 0.693 or x = \ln 3 \approx 1.099. With u = e^x, u^2 - 5u + 6 = (u - 2)(u - 3) = 0, so e^x = 2 or e^x = 3; both are positive, so both give solutions.
After about 7.82 years, during the 8th year. \displaystyle (1.021)^t = \frac{100}{85}, so \displaystyle t = \frac{\ln \frac{20}{17}}{\ln 1.021} \approx 7.82.
The balance is \displaystyle 2500\left(1 + \frac{0.045}{4}\right)^{4t} = 2500(1.01125)^{4t}. Doubling means (1.01125)^{4t} = 2, so \displaystyle 4t = \frac{\ln 2}{\ln 1.01125} \approx 61.96 and t \approx 15.49 years. Interest is added quarterly, so the balance first reaches 5{,}000 dollars after 62 quarters, which is 15.5 years (about 5{,}002.29 dollars; after 61 quarters it is about 4{,}946.64).
After about 10.22 years. \displaystyle e^{0.05t} = \frac{5}{3}, so \displaystyle t = \frac{\ln \frac{5}{3}}{0.05} \approx 10.22.
After about 8.59 years. \displaystyle (0.88)^t = \frac{1}{3}, so \displaystyle t = \frac{\ln \frac{1}{3}}{\ln 0.88} \approx 8.59.
After about 24.76 years. In thousands, 40(1.03)^t = 65(1.01)^t; dividing, \displaystyle \left(\frac{1.03}{1.01}\right)^t = \frac{65}{40} = 1.625, so \displaystyle t = \frac{\ln 1.625}{\ln \frac{1.03}{1.01}} \approx 24.76.
\displaystyle k = \frac{\ln 0.84}{6} \approx -0.02906, half-life about 23.85 hours. From 60e^{6k} = 50.4, e^{6k} = 0.84. Then \displaystyle e^{kT} = \frac{1}{2} gives \displaystyle T = \frac{-\ln 2}{k} = \frac{6\ln 2}{\ln \frac{1}{0.84}} \approx 23.85.
With a = b = 1: \log(1 + 1) = \log 2 \approx 0.301, but \log 1 + \log 1 = 0. The left side 5 \cdot 3^x + 8 is a sum, and nothing turns the logarithm of a sum into something simpler, so the 8 must be subtracted first. The factor 5 multiplies the power, and the product law splits a product: \log\left(5 \cdot 3^x\right) = \log 5 + x\log 3, which can be solved.
Since c = 10^{\log c}, taking natural logarithms gives \ln c = \log c \cdot \ln 10, and in the same way \ln b = \log b \cdot \ln 10. Dividing, the factor \ln 10 cancels: \displaystyle \frac{\ln c}{\ln b} = \frac{\log c}{\log b}. (\log b \neq 0 because b \neq 1.)
If c \leq 0 there is no solution, since b^x > 0 for every real x. If c > 0, then \displaystyle x = \frac{\log c}{\log b} is a solution, as the steps can be reversed. It is the only one, because b^x is always increasing (when b > 1) or always decreasing (when 0 < b < 1), so it never takes the same value twice.
No. \displaystyle \frac{\ln 40}{\ln 7} \approx 1.896 and \displaystyle \ln \frac{40}{7} \approx 1.743. The first solves 7^x = 40; the second is \ln 40 - \ln 7, a difference of logarithms, not a quotient.