Question

(1.07)^4x = 5/2 Solve for x.

Answer 1

\[1.07^{4x}\]
or
\[1.07^{4}x\]
???

Answer 2

^(4x)

Answer 3

ok
first step it to take the log of both sides
log ((1.07)^(4x)) = log (5/2)
what's the next step? remember: log(a)^b = b*log(a)

Answer 4

slight correction
remember: log(a^b) = b*log(a)

Answer 5

\(\bf 1.07^{4x}=\cfrac{5}{2}\\ \quad \\
\textit{log cancellation of }\quad log_aa^x = x\\ \quad \\
log_{1.07}(1.07^{4x})=log_{1.07}\left(\cfrac{5}{2}\right)\implies 4x = log_{1.07}\left(\cfrac{5}{2}\right)\\ \quad \\
\textit{change of base rule}\quad log_ab=\cfrac{log_ca}{log_cb}\\ \quad \\

4x = \cfrac{log_{10}\left(\frac{5}{2}\right)}{log_{10}(1.07)}\)

Answer 6

\(\bf \textit{change of base rule}\quad log_ab=\cfrac{log_cb}{log_ca}\) to be exact =)