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.
Thanks in advance!
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.
Thanks in advance!
\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*}
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} $