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

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Answer 1 · 2015-12-06T20:09:41

$x = 5$ $x=5$

Using laws of exponents we may write this equation as

$3^{2 x} \cdot 3^{5} = 3^{3 x}$ $3^(2x)*3^5=3^(3x)$

$therefore3^5=3^(3x-2x)$

$therefore 3^5=3^x$ and hence $x=5$

How do you solve 3^(2x+5)=27^x32x+5=27x3^(2x+5)=27^x?