Question

How do you solve $7\times 3^{\mathrm{t}}=5\times 2^{\mathrm{t}}$?

Answer 1

Determine to find $7\times 3^{\mathrm{t}}=5\times 2^{\mathrm{t}}$:

Take $\log$ on both sides
$\begin{array}{rcl}\log (7\times 3^{\mathrm{t}})& =& \log (5\times 2^{\mathrm{t}})\\ \begin{array}{c}\Rightarrow \end{array}\log \left(7\right)+\log \left(3^{\mathrm{t}}\right)& =& \log \left(5\right)+\log \left(2^{\mathrm{t}}\right)\\ \begin{array}{c}\Rightarrow \end{array}\log \left(7\right)+\mathrm{t}\log \left(3\right)& =& \log \left(5\right)+\mathrm{t}\log \left(2\right)\\ \begin{array}{c}\Rightarrow \end{array}\log \left(7\right)-\log \left(5\right)& =& \mathrm{t}\log \left(2\right)-\mathrm{t}\log \left(3\right)\end{array}$

Continue the evaluation as given below

$\begin{array}{rcl}\begin{array}{c}\Rightarrow \end{array}\log \left(\frac{7}{5}\right)& =& \mathrm{t}\left(\log \right(2)-\log (3\left)\right)\left[\log a-\log =\log \left(\frac{a}{b}\right)\right]\\ \begin{array}{c}\Rightarrow \end{array}\log \left(\frac{7}{5}\right)& =& \mathrm{t}\left(\log \left(\frac{2}{3}\right)\right)\\ \begin{array}{c}\Rightarrow \end{array}\frac{\log \left({\displaystyle \frac{7}{5}}\right)}{\log \left({\displaystyle \frac{2}{3}}\right)}& =& \mathrm{t}\\ \begin{array}{c}\Rightarrow \end{array}\mathbf{t}& \mathbf{=}& \mathbf{-}\mathbf{0}\mathbf{.}\mathbf{8298}\end{array}$
Therefore, the value of $7\times 3^{\mathrm{t}}=5\times 2^{\mathrm{t}}$ is $\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{8298}$.