# How do you solve log_3 (4x-5)=5?

Mar 15, 2018

I got $x = 62$

#### Explanation:

Let us use the definition of log to write:

$4 x - 5 = {3}^{5}$

rearrange:

$4 x = 243 + 5$

$x = \frac{248}{4} = 62$

NB: definition of log
${\log}_{b} a = x$
so that:
$a = {b}^{x}$

Mar 15, 2018

The solution is $x = 62$.

#### Explanation:

Convert the equation to exponential form:

${\log}_{\textcolor{red}{a}} \left(\textcolor{g r e e n}{y}\right) = \textcolor{b l u e}{x} q \quad q \quad \iff q \quad q \quad {\textcolor{red}{a}}^{\textcolor{b l u e}{x}} = \textcolor{g r e e n}{y}$

Here's the actual equation:

$\textcolor{w h i t e}{\implies} {\log}_{\textcolor{red}{3}} \left(\textcolor{g r e e n}{4 x - 5}\right) = \textcolor{b l u e}{5}$

$\implies {\textcolor{red}{3}}^{\textcolor{b l u e}{5}} = \textcolor{g r e e n}{4 x - 5}$

$\textcolor{w h i t e}{\implies} 243 = 4 x - 5$

$\textcolor{w h i t e}{\implies} 248 = 4 x$

$\textcolor{w h i t e}{\implies} 62 = x$

$\textcolor{w h i t e}{\implies} x = 62$

That's the answer. Hope this helped!

Mar 15, 2018

Given: ${\log}_{3} \left(4 x - 5\right) = 5$

Make both sides the exponent of the base, 3:

${3}^{{\log}_{3} \left(4 x - 5\right)} = {3}^{5}$

The left side simplifies to the argument of the logarithm and the right side is computed with a calculator:

$4 x - 5 = 243$

Solve for x:

$4 x = 248$

$x = 62$