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?
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?
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.)
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.
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$.
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).