# How do we solve 6^3t-1=5 for t?

Jun 5, 2016

$t = \frac{1}{3} {\log}_{6} \left(30\right)$

#### Explanation:

In the absence of proper formatting I believe it is

${6}^{3 t - 1} = 5$. If so

${\log}_{6} 5 = 3 t - 1$ or

$3 t = {\log}_{6} 5 + 1 = {\log}_{6} 5 + {\log}_{6} 6 = {\log}_{6} \left(5 \times 6\right)$

$3 t = {\log}_{6} \left(30\right)$

Hence, $t = \frac{1}{3} {\log}_{6} \left(30\right)$

Jun 5, 2016

For ${6}^{3 t} - 1 = 5$, the answer is $t = \frac{1}{3}$.
For ${6}^{3 t - 1} = 5$, the answer is $t = \frac{1 + \log \frac{5}{\log} 6}{3}$.

#### Explanation:

If it is ${6}^{3 t} - 1 = 5$ then ${6}^{3 t} = 6$.

So, $3 t = 1 \mathmr{and} t = \frac{1}{3}$.

For ${6}^{3 t - 1} = 5$, equate logarithms.

$\left(3 t - 1\right) \log 6 = \log 5$. Solving,

$t = \frac{1 + \log \frac{5}{\log} 6}{3}$.