What is the value of $x$ $x$ if $2^{3} (3^{2 x + 3}) = 6 {(2)}^{2}$ $2^3(3^(2x + 3)) = 6(2)^2$ ?

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Answer 1 · 2016-12-21T23:55:52

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

Isolate the exponent.

$3^{2 x + 3} = \frac{24}{8}$ $3^(2x + 3) = 24/8$

$3^{2 x + 3} = 3$ $3^(2x+ 3) = 3$

We are in common bases--logarithms are not necessary.

$2 x + 3 = 1$ $2x + 3 = 1$

$2 x = - 2$ $2x = -2$

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

Check back in the original equation:

$8 (3^{2 (- 1) + 3}) =^{?} 24$ $8(3^(2(-1) + 3)) =^? 24$

$8 (3^{1}) = 24 \sqrt$ $8(3^1) = 24" "color(green)(√)$

Practice Exercises

a) $3^{2 x + 5} = 9^{x - 2}$ $3^(2x+ 5) = 9^(x - 2)$

b) $\frac{2^{5 x + 1}}{16^{3 x}} = 8^{4 x + 2}$ $2^(5x + 1)/16^(3x) = 8^(4x + 2)$

Hopefully this helps!

What is the value of xx if 23(32x+3)=6(2)22^3(3^(2x + 3)) = 6(2)^2?