Question

Solve for t.

Answer 1

\[3^{t+8}=9^{2t}\]

Answer 2

I assume you're covering logarithms?

Answer 3

Yep.

Answer 4

\[3^{t+8}=9^{2t}=\left\{ \left( 3 \right)^2 \right\}^{2t}=3^{4t}\]
t+8=4t
t=?

Answer 5

something to keep in mind
log cancellation rules of \(\bf \large log_{\color{red}{ a}}{\color{red}{ a}}^x\implies x\qquad \qquad
{\color{red}{ a}}^{log_{\color{red}{ a}}x}=x\)

Answer 6

I'm not sure how you got there sshayer.
However, I converted it to
t+8log3=2tlog9

Answer 7

Oh, I see how you got there sshayer. But do you know how to do it with the method where you add log to the equation?

Answer 8

\(\large {
3^{t+8}=9^{2t}\implies 3^{t+8}=(3^2)^{2t}\implies 3^{t+8}=3^{4t}
\\ \quad \\
\textit{using the log cancellation rule above}
\\ \quad \\
log_{\color{red}{ 3}}\left( {\color{red}{ 3}}^{t+8} \right)=log_{\color{red}{ 3}}\left( {\color{red}{ 3}}^{4t} \right)\implies t+8=4t
}\)

Answer 9

solving by log
\[3 ^{t+8}=9^{2t}=\left\{ \left( 3 \right) ^2\right\}^{2t}=3^{4t}\]
take log of bothsides
\[\left( t+8 \right)\log 3=4tlog3\]
divide both sides log3
t+8=4t
t=?

Answer 10

\[\log~a^x=x~\log~a\]

Answer 11

I get it, thank you!

Answer 12

yw

Answer 13

you could also look at it, the longer way
\(\bf log_{\color{red}{ 3}}\left( {\color{red}{ 3}}^{t+8} \right)=log_{\color{red}{ 3}}\left( {\color{red}{ 3}}^{4t} \right)\implies (t+8)(log_3(3))=(4t)(log_3(3))
\\ \quad \\
t+8=\cfrac{4t\cancel{(log_3(3))}}{\cancel{(log_3(3))}}\)