Question

How do you solve for x?tle=" log_{6}(36) = 5x + 6" alt=" log_{6}(36) = 5x + 6" align="absmiddle" class="latex-formula">
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Answer 1

The value of x in the given equation is $-\frac{4}{5}$

From the question,

We are to solve the given equation

The given equation is

$log_{6}(36) = 5x +6$

This becomes

$log_{6}(6^{2} ) = 5x +6$

From one of the laws of logarithms, we have that

$log_{x}(x^{2} ) = 2log_{x}(x)$

and

$log_{x}(x) = 1$

∴ $log_{6}(6^{2} ) = 5x +6$ becomes

$2log_{6}(6) = 5x +6$

and

$2(1)= 5x +6$

$2 = 5x +6$

Now, subtract 6 from both sides

$2-6 = 5x +6-6$

$-4 = 5x$

∴ $x = -\frac{4}{5}$

Hence, the value of x in the given equation is $-\frac{4}{5}$

Learn more on solving equations here: brainly.com/question/11802986

Answer 2

$\huge \rm༆ Answer ༄$

The value of x is ~

• $-0.8$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

Let's solve for x ~

• $\sf \: log_{6}(36) = 5x + 6$

• $2 = 5x + 6$

• $5x = 2 - 6$

• $x = - 4 \div 5$

• $x = - 0.8$

I hope it helped ~

$꧁ \: \large \frak{Eternal \: Being } \: ꧂$