How do you solve $36^{x - 9} = 6^{2 x}$ $36^(x-9)=6^(2x)$ ?

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Answer 1 · 2016-10-07T11:08:38

The equation has no solutions

We can solve exponential equation if the basis is the same, since

$a^{x} = a^{y} \Leftrightarrow x = y$ $a^x = a^y \iff x=y$

To bring your equation to this form, we simply need to observe that $36 = 6^{2}$ $36=6^2$ , and thus we have

$36^{x - 9} = {(6^{2})}^{x - 9}$ $36^{x-9} = (6^2)^{x-9}$

Now use the rule ${(a^{b})}^{c} = a^{b \cdot c}$ $(a^b)^c = a^{b*c}$ to obtain

${(6^{2})}^{x - 9} = 6^{2 (x - 9)} = 6^{2 x - 18}$ $(6^2)^{x-9} = 6^{2(x-9)}=6^{2x-18}$

So now the equation looks like

$6^{2 x - 18} = 6^{2 x}$ $6^{2x-18} = 6^{2x}$

which would be true only if

$2 x - 18 = 2 x$ $2x-18=2x$ , which is clearly impossible

How do you solve 36x−9=62x36^(x-9)=6^(2x)?