How do you solve $3^{x - 1} = \frac{27}{3^{x}}$ $3^(x-1)= 27/3^x$ ?

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Answer 1 · 2016-06-07T07:40:45

We can use the laws of exponents and simplify to get the answer.
Answer : $x = 2$ $x = 2$

$3^{x - 1} = \frac{27}{3^{x}}$ $3^(x-1) = 27/ (3^x)$

$3^{x - 1} = \frac{3^{3}}{3^{x}}$ $3^(x-1) = 3^3/3^x$

$3^{x - 1} = 3^{3 - x}$ $3^(x-1) = 3^(3-x)$

As the bases are equal, we can equate the exponents.

$x - 1 = 3 - x$ $x - 1 = 3 - x$

$x + x = 3 + 1$ $x + x = 3 + 1$

$2 x = 4$ $2x = 4$

$x = 2$ $x = 2$

How do you solve 3^(x-1)= 27/3^x3x−1=273x3^(x-1)= 27/3^x?