# How do I find the solution for these exponential equations?

I'm trying to solve these two exponential equations to four decimal places...

$4^{5x − 4} = 8$

$(1/8)^x = 85$

But I keep coming up with the wrong answers...help?

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What have you been trying? –  Austin Mohr Oct 10 '12 at 0:00
Well, for the second one I did this...log[(1/8)^x] = 85, then (1/8)logx = log85, then 1/8 = 85/x, but I get stuck. –  Brandt Oct 10 '12 at 0:25
@Brandt: You should have $x \log(1/8) = \log(85)$, not $(1/8) \log(x) = \log(85)$. –  mike4ty4 Oct 10 '12 at 0:35

First problem: $$4^{5x -4} = 8 = 2^3 = 2^{2(5x-4)} = 2^{10x -8} = 2^3\\ \implies 10x - 8 = 3 \\ \implies10x = 11\\ \implies x = 11/10.$$

Second problem:

$$(1/8)^x = 85 = 5\cdot 17 \\ \implies x\log (1/8) = \log (5\cdot17) = \log 5 + \log 17\\=x \log (2^{-3}) =-3x\log 2\\ \implies x\log 2 = -(1/3)(\log 5 + \log 17)\\ \implies x = - \frac{\log 5 + \log 17}{3 \log 2}$$

Now, I assume you can approximate the answer as much as you want using a calculator or a computer. Wolfram says $x\approx -2.13646$.

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You can use logarithms to solve both of these equations. Taking the log of both sides results in equations that are simple to solve for $x$.

For example, the first equation results in $$(5x-4)\log 4 = \log 8.$$ Thus, $$x = \frac{\frac{\log8}{\log4}+ 4}{5},$$ which can be found to as many decimal places as you want.

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But $\log 8 = 3\log2$, and $\log 4 = 2\log 2$. Then $x = 11/10$. There's no need to compute it by approximations; it is a rational number. –  Ian Mateus Oct 10 '12 at 0:25

For the second problem, take the logarithm of both sides:log (1/8)^x = log 85. Then apply the power rule for logs: x log (1/8) = log 85. Note: log (1/8) = log 1 - log 8 because of the quotient rule for logs and the log 1 = 0. Therefore, (-log 8)x = log 85. Dividing both sides by -log 8, gives x = -(log 85/log 8).

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You can just take the $\log$ of both sides with base equal to the base of the given exponentiation. Then the rest is easy. (And if you need it, just remember that $\log$ to base $b$ is $\log_b(x) = \frac{\log(x)}{\log(b)}$, where the other $\log$ is in a base you can compute logs to.)

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