Solve for $x$.
What if the bases cannot be made the same? For example, $2^x=7$?
$2^x=7$
$\textcolor{#17a2b8}{\log}{2^x}=\textcolor{#17a2b8}{\log}{7}$
$\log{2}=\log{7}$
$x=$
$2^x=7$
$\textcolor{#17a2b8}{\log_2}{2^x}=\textcolor{#17a2b8}{\log_2}{7}$
${\log_2}{2}=\log_2{7}$
$=$
$2^x=7$
$\textcolor{#17a2b8}{\ln}{2^x}=\textcolor{#17a2b8}{\ln}{7}$
$\ln{2}=\ln{7}$
$x=$
They are are equivalent (you can check on your calculator)
Solve for $x$.
$5\left(4^{2x}\right)=15$
$4^{2x}=$
$\textcolor{#17a2b8}{\log}{4^{2x}}=\textcolor{#17a2b8}{\log}{3}$
$\log{4}=\log{3}$
$2x=$
$x=$
$\textcolor{#17a2b8}{\log_4}{4^{2x}}=\textcolor{#17a2b8}{\log_4}{3}$
$\log_4{4}=\log_4{3}$
$2x=$
$x=$
$\textcolor{#17a2b8}{\ln}{4^{2x}}=\textcolor{#17a2b8}{\ln}{3}$
$\ln{4}=\ln{3}$
$2x=$
$x=$
Solve for $x$.
$2e^{4x-1}-3=7$
$e^{4x-1}=$
$\textcolor{#17a2b8}{\log}{e^{4x-1}}=\textcolor{#17a2b8}{\log}{5}$
$\log{e}=\log{5}$
$4x-1=$
$x=\;$ / 4
$\textcolor{#17a2b8}{\log_a}{e^{4x-1}}=\textcolor{#17a2b8}{\log_a}{5}$
$\log_a{e}=\log_a{5}$
$4x-1=$
$x=\;$ / 4
$\textcolor{#17a2b8}{\ln}{e^{4x-1}}=\textcolor{#17a2b8}{\ln}{5}$
$\ln{e}=\ln{5}$
$4x-1=$
$x=\;$ / 4