How do you solve $81^{x} = 27^{x + 2}$ $81^x = 27^(x + 2)$ ?

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Answer 1 · 2015-07-22T19:25:42

More simply, $x = 6$ $x=6$

(logarithms are unnecessary in this case).

$81 = 3^{4}$ $81=3^{4}$ and $27 = 3^{3}$ $27=3^{3}$ , so this equation can also be written as

$3^{4 x} = 3^{3 (x + 2)} = 3^{3 x + 6}$ $3^{4x}=3^{3(x+2)}=3^{3x+6}$ .

Since the bases are now the same, we can equate exponents to get

$4 x = 3 x + 6$ $4x=3x+6$ so that

$x = 6$ $x=6$

(technically this equating of exponents requires the fact that exponential functions (with base not equal to 1) are one-to-one functions).

How do you solve 81x=27x+281^x = 27^(x + 2)?