How do you solve $5 = 6^{3 t - 1}$ $5=6^(3t-1)$ ?

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Answer 1 · 2016-08-02T18:38:55

$t = \frac{1}{3} (\frac{log 5}{log 6} + 1)$ $t = 1/3((log 5)/(log 6)+1)$

Taking logs of both sides, we have:

$log 5 = log (6^{3 t - 1}) = (3 t - 1) log 6$ $log 5 = log (6^(3t-1)) = (3t-1) log 6$

Divide both ends by $log 6$ $log 6$ and transpose to get:

$3 t - 1 = \frac{log 5}{log 6}$ $3t - 1 = (log 5)/(log 6)$

Add $1$ $1$ to both sides to get:

$3 t = \frac{log 5}{log 6} + 1$ $3t = (log 5)/(log 6) + 1$

Divide both sides by $3$ $3$ to get:

$t = \frac{1}{3} (\frac{log 5}{log 6} + 1)$ $t = 1/3((log 5)/(log 6)+1)$

How do you solve 5=6^(3t-1)5=63t−15=6^(3t-1) ?