How do you solve $3^{5 x} = 27^{2 x + 1}$ $3 ^(5x) = 27^(2x + 1)$ ?

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Answer 1 · 2016-04-14T03:46:09

Rearrange and take the logarithms to the base 3 of both sides, to yield $x = - 3$ $x=-3$

Rewrite the right hand side:

$27^{2 x + 1} = 3^{3 (2 x + 1)} = 3^{6 x + 3}$ $27^(2x+1)=3^(3(2x+1))=3^(6x+3)$

Now the overall equation is:

$3^{5 x} = 3^{6 x + 3}$ $3^(5x)=3^(6x+3)$

Taking the log to the base 3 of both sides:

$5 x = 6 x + 3$ $5x=6x+3$

$x = - 3$ $x=-3$

How do you solve 35x=272x+13 ^(5x) = 27^(2x + 1)?