Certain exponential equations arise that can be easily solved without resorting to logarithms.
The key to their solution lies in the recognition of various powers of integers. for example recognizing that $81=3^4$81=34, or $125=5^3$125=53 or $256=2^8=4^4=16^2$256=28=44=162.
With a good understanding of the rules of indices, in particular rules like $a^{-1}=\frac{1}{a}$a−1=1a and $a^{\frac{1}{2}}=\sqrt{a}$a12=√a, and more generally $a^{-m}=\frac{1}{a^m}$a−m=1am and $a^{\frac{1}{m}}=\sqrt[m]{a}$a1m=^{m}√a, solutions can be readily found.
Here is a list of typical examples. Note that in each case it is possible to re-express certain numbers in the equation as powers. If this were not the case, then we would resort to a different strategy - one that involved the laws of logarithms.
If $a^m=a^n$am=an, then $m=n$m=n, so we would like to get both sides of the equation with the same base.
Solve $7^p=343$7p=343
Think: Can $343$343 be rewritten as an exponent with base $7$7?
Do:
$7^p$7p | $=$= | $343$343 |
$7^p$7p | $=$= | $7^3$73 |
$\therefore$∴ $p$p | $=$= | $3$3 |
Solve $2^{1-2x}=1024$21−2x=1024
Think: Can $1024$1024 be rewritten as an exponent with base $2$2?
Do:
$2^{1-2x}$21−2x | $=$= | $1024$1024 |
$2^{1-2x}$21−2x | $=$= | $2^{10}$210 |
$1-2x$1−2x | $=$= | $10$10 |
$2x$2x | $=$= | $-9$−9 |
$\therefore$∴ $x$x | $=$= | $-4\frac{1}{2}$−412 |
Solve $\left(3^2\right)^{x+1}=\frac{1}{81}$(32)x+1=181
Think: Can the fraction be rewritten as an exponent with base $3$3?
Do:
$\left(3^2\right)^{x+1}$(32)x+1 | $=$= | $\frac{1}{81}$181 |
$\left(3^2\right)^{x+1}$(32)x+1 | $=$= | $\frac{1}{3^4}$134 |
$3^{2x+2}$32x+2 | $=$= | $3^{-4}$3−4 |
$2x+2$2x+2 | $=$= | $-4$−4 |
$2x$2x | $=$= | $-6$−6 |
$\therefore$∴ $x$x | $=$= | $-3$−3 |
Solve $2^{2x}-20\left(2^x\right)+64=0$22x−20(2x)+64=0
Think: The equation looks like a quadratic equation. Do not forget to factor if needed.
Do:
$2^{2x}-20\left(2^x\right)+64$22x−20(2x)+64 | $=$= | $0$0 |
$\left(2^x\right)^2-20\left(2^x\right)+64$(2x)2−20(2x)+64 | $=$= | $0$0 |
Setting $u$u | $=$= | $2^x$2x |
$u^2-20u+64$u2−20u+64 | $=$= | $0$0 |
$\left(u-16\right)\left(u-4\right)$(u−16)(u−4) | $=$= | $0$0 |
$\therefore$∴ $u$u | $=$= | $4,16$4,16 |
Since $u=2^x$u=2x, we have $2^x=4$2x=4 and $2^x=16$2x=16, and thus $x=2$x=2 and $x=4$x=4.
Solve $8^y=\sqrt[3]{32}$8y=^{3}√32
Think: Change the square root into an exponent. Then, rewrite the both sides so they have the same base.
Do:
$8^y$8y | $=$= | $\sqrt[3]{32}$^{3}√32 |
$\left(2^3\right)^y$(23)y | $=$= | $\left(2^5\right)^{\frac{1}{3}}$(25)13 |
$2^{3y}$23y | $=$= | $2^{\frac{5}{3}}$253 |
$3y$3y | $=$= | $\frac{5}{3}$53 |
$\therefore$∴ $y$y | $=$= | $\frac{5}{9}$59 |
Solve $49\left(7^x\right)=2401\left(\sqrt{7}\right)^{2-4x}$49(7x)=2401(√7)2−4x
Think: Find a common base for all of the numbers. Then change each to have the common base. We must be attention of the laws of exponents.
Do:
$49\left(7^x\right)$49(7x) | $=$= | $2401\left(\sqrt{7}\right)^{2-4x}$2401(√7)2−4x |
$7^2\times7^x$72×7x | $=$= | $7^4\left(7^{\frac{1}{2}}\right)^{2-4x}$74(712)2−4x |
$7^{2+x}$72+x | $=$= | $7^4\times7^{1-2x}$74×71−2x |
$7^{2+x}$72+x | $=$= | $7^{5-2x}$75−2x |
$2+x$2+x | $=$= | $5-2x$5−2x |
$\therefore$∴ $y$y | $=$= | $1$1 |
Solve the equation $\left(2^2\right)^{x+7}=2^3$(22)x+7=23 for $x$x.
Consider the equation
$\left(2^x\right)^2-9\times2^x+8=0$(2x)2−9×2x+8=0
The equation can be reduced to a quadratic equation by using a certain substitution.
By filling in the gaps, determine the correct substitution that would reduce the equation to a quadratic.
Let $m=\left(\editable{}\right)^{\editable{}}$m=()
Solve the equation for $x$x by using the substitution $m=2^x$m=2x.
Solve the equation $25^{x+1}=125^{3x-4}$25x+1=1253x−4.
We have seen that the value of a logarithm such as $\log_ba$logba is "the power to which $b$b must be raised in order to give $a$a".
That is, for a logarithmic equation of the form $\log_ba=c$logba=c, the value of $c$c is the power of $b$b that will give $a$a.
Algebraically, this means that the equation $\log_ba=c$logba=c is equivalent to the exponential equation $b^c=a$bc=a.
We can make use of this key relationship between logarithms and exponentials in order to solve logarithmic equations. First, rewrite the logarithmic equation in terms of exponentials, then use the properties of exponentials that we are familiar with to solve the equation.
Solve the equation $\log_2x=7$log2x=7 for $x$x.
Think: This equation is in the form $\log_ba=c$logba=c, so we can start by rewriting it without logarithms.
Do: $\log_2x=7$log2x=7 is equivalent to $x=2^7$x=27, and so the solution is $x=128$x=128.
A logarithmic equation of the form $\log_ba=c$logba=c is equivalent to the exponential equation $b^c=a$bc=a.
We can use this to solve some equations by first rewriting them without logarithms.
One method to solve equations of the form $a^x=b$ax=b is to take the logarithm of both sides. This will let us rearrange the equation to get $x$x in terms of $\log a$loga and $\log b$logb.
Solve $12^x=30$12x=30 for $x$x as both an exact value and to three decimal places.
Think: This is an equation of the form $a^x=b$ax=b. So we can solve it by taking the logarithm of both sides.
Do:
$12^x$12x | $=$= | $30$30 | |
$\log12^x$log12x | $=$= | $\log30$log30 | (Take the logarithm of both sides) |
$x\log12$xlog12 | $=$= | $\log30$log30 | (Using the identity $\log A^B=B\log A$logAB=BlogA) |
$x$x | $=$= | $\frac{\log30}{\log12}$log30log12 | |
$\approx$≈ | $1.369$1.369 | (To three decimal places) |
Reflect: Notice that we didn't specify the base of the logarithm. It doesn't actually matter because of the identity $\log_AB=\frac{\log B}{\log A}$logAB=logBlogA. In fact, we can skip a step by taking the logarithm of base $12$12 here. However, it is easier to use logarithms of base $10$10 when evaluating the solution with a calculator.
Solve $\log_45x=3$log45x=3 for $x$x.
Solve $\log_{10}\left(3x+982\right)=3$log10(3x+982)=3 for $x$x.
Solve $11\log_5\left(x-12\right)=33$11log5(x−12)=33 for $x$x.
Consider the equation $5^x=\frac{1}{11}$5x=111.
Rearrange the equation into the form $x=\frac{\log A}{\log B}$x=logAlogB.
Evaluate $x$x to three decimal places.
We learned how to evaluate logarithmic expressions in terms of numbers, e.g. $\log_39=2$log39=2. Now, let's have a look at how to use the laws of logs in algebraic equations to solve equations.
Solve: $\log_22+\log_2\left(3x-5\right)=3$log22+log2(3x−5)=3
Think: How many $\log$log are there and what operation is being preformed? There are $2$2 $\log$log and the operation being preformed is addition. Therefore, we will use the product rule to create one $\log$log.
Do:
$\log_22+\log_2\left(3x-5\right)$log22+log2(3x−5) | $=$= | $3$3 |
(Given) |
$\log_22\left(3x-5\right)$log22(3x−5) | $=$= | $3$3 |
(Use the product rule) |
$\log_2\left(6x-10\right)$log2(6x−10) | $=$= | $3$3 |
(Distribute) |
$2^3$23 | $=$= | $6x-10$6x−10 |
(Rewrite as an exponent) |
$8$8 | $=$= | $6x-10$6x−10 |
(Calculate $2^3$23) |
$18$18 | $=$= | $6x$6x |
(Add $10$10 to both sides) |
$3$3 | $=$= | $x$x |
(Divide by $6$6 to both sides) |
Answer: $x=3$x=3
Solve: $2\ln x+3=7$2lnx+3=7
Think: We need to isolate the natural logarithm.
Do:
$2\ln x+3$2lnx+3 | $=$= | $7$7 |
(Given) |
$2\ln x$2lnx | $=$= | $4$4 |
(Subtract $3$3 from both sides) |
$\ln x$lnx | $=$= | $2$2 |
(Divide by $2$2 to both sides) |
$x$x | $=$= | $e^2$e2 |
(Rewrite) |
Solve: $\log\left(3x-2\right)-\log2=\log\left(x+4\right)$log(3x−2)−log2=log(x+4)
Think: How many logarithms are in the equation? $3$3 logs, with two on the same side with subtraction separating them. Therefore we will have to use the quotient rule to condense the number of logs.
Do:
$\log\left(3x-2\right)-\log2$log(3x−2)−log2 | $=$= | $\log\left(x+4\right)$log(x+4) |
(Given) |
$\log\left(\frac{3x-2}{2}\right)$log(3x−22) | $=$= | $\log\left(x+4\right)$log(x+4) |
(Use the quotient rule) |
$\frac{3x-2}{2}$3x−22 | $=$= | $x+4$x+4 |
(Use the one-to-one property) |
$3x-2$3x−2 | $=$= | $2x+8$2x+8 |
(Multiply $2$2 to both sides) |
$x-2$x−2 | $=$= | $8$8 |
(Subtract $2x$2x to both sides) |
$x$x | $=$= | $10$10 |
(Add $2$2 to both sides) |
Reflect: Check your result by substituting $x=10$x=10 into the given equation:
$\log\left(3x-2\right)-\log2$log(3x−2)−log2 | $=$= | $\log\left(x+4\right)$log(x+4) |
$\log\left(3\times10-2\right)-\log2$log(3×10−2)−log2 | $=$= | $\log\left(10+4\right)$log(10+4) |
$\log28-\log2$log28−log2 | $=$= | $\log14$log14 |
$\log\frac{28}{2}$log282 | $=$= | $\log14$log14 |
$\log14$log14 | $=$= | $\log14$log14 |
Solve $2^x=5$2x=5 for $x$x.
Give your answer to 2 decimal places if necessary.
Solve $\log_7y=5$log7y=5 for $y$y.
Solve $\log_{10}x-\log_{10}38=\log_{10}37$log10x−log1038=log1037 for $x$x.
For exponential models, express as a logarithm the solution to ab^(ct) = d wherea, c, and d are numbers and the base b is 2, 10, or e; evaluate the logarithm using technology.