Solve .
Could someone help explain how to solve this? Cheers
Hi,If you are supposed to know what $\displaystyle \sinh$ is you can claim that
$\displaystyle 6^x-6^{-x}=\frac{15}{4} \Longleftrightarrow \mathrm{e}^{x\ln 6}-\mathrm{e}^{-x\ln 6}=\frac{15}{4} \Longleftrightarrow \sinh(x\ln6)=\frac{15}{8}$.
And this equation can easily be solved using the inverse function of $\displaystyle \sinh$.
Follow Moo's advice.
$\displaystyle \begin{gathered}
4\left( {6^{2x} } \right) - 15\left( {6^x } \right) - 4 = 0 \hfill \\
\left[ {4\left( {6^x } \right) + 1} \right]\left[ {\left( {6^x } \right) - 4} \right] = 0 \hfill \\ \left( {6^x } \right) \ne \frac{{ - 1}}
{4}\,\& \,\left( {6^x } \right) = 4 \hfill \\
\end{gathered} $
Can you finish?