Solving the equation $3^{5x-2}=8^{8x-9}$

I'm trying to solve the equation $$3^{5x-2}=8^{8x-9}.$$

I'm assuming I need to do some work with logarithms, but I don't know what to do.

• Is this homework? What laws for lograithms do you know? For the formatting: Use curly brackets for exponents ... Sep 17 '11 at 18:24
• Hint: Let us write $\log x$ for the logarithm of $x$ to your favourite base ($10$, $e$, $2$, it doesn't matter). We have in general $\log(a^b)=b\log a$. Sep 17 '11 at 18:44
• Did you look at the logarithm laws - did you try anything?
Sep 17 '11 at 18:52
• this homework isn't going to be marked, it's simply review for my calculus course. I know all of the main log laws (subtracting log, addition of log, changing base, and the a*log(x) = log(x^a) Sep 17 '11 at 19:11

\begin{align*} \ln 3^{5x-2} &= \ln 8^{8x-9}\\ \iff \ln 3 \cdot (5x-2) &= \ln 8 \cdot (8x-9) \\ \iff 8\ln 8 \cdot x-5\ln 3 \cdot x &=-2\ln 3+9\ln 8\\ \iff x(8\ln 8-5\ln 3)&=-2\ln 3+9\ln 8\\ \iff x&=\frac{-2\ln 3+9\ln 8}{8\ln 8-5\ln 3}. \end{align*}

• great answer @Andres ,he needs only excel and using function ln he can calculate everything Sep 17 '11 at 18:44
• thanks a bunch. I was trying this approach, but I used a base of 3 or 8 so I could use one of the other laws and I just ended up confusing myself. This is quite straightforward :) Sep 17 '11 at 19:16

Yes, you could apply log's to both sides and then solve the linear equations for $\:x$.

Alternatively, dually, you can trade off knowledge of logs for exponents. Namely rewrite it as

$$\dfrac{8^{\:9}}{3^{\:2}}\ =\ \bigg(\dfrac{8^{\:8}}{3^{\:5}}\bigg)^x$$

$$\Rightarrow\quad x\ =\ \dfrac{\log(8^9/3^2)}{\log(8^8/3^5)}\quad\ \$$

Using base 3 or base 8 will work just as well: \begin{align} 3^{5x-2} & =8^{8x-9} \\ \\ 5x-2 & = \log_3 \left(8^{8x-9}\right) = (8x-9)\log_3 8 = 8(\log_3 8)x -9\log_3 8 \\ \\ 5x - 8(\log_3 8)x & = 2 - 9\log_3 8 \\ \\ (5 - 8\log_3 8)x & = 2 - 9\log_3 8 \\ \\ x & = \frac{2 - 9\log_3 8}{5 - 8\log_3 8}. \end{align}