Solve the following exponential equations. 7) 3^2x-2=27 8) (1^x)/(8)=4^x+7 7 __ 8. __

7) To solve the equation $3^{2x-2}=27$, we can rewrite 27 as $3^3$ and then use the property of exponents that states $a^m = a^n$ implies $m = n$.<br /><br />$3^{2x-2} = 3^3$<br /><br />$2x - 2 = 3$<br /><br />$2x = 5$<br /><br />$x = \frac{5}{2}$<br /><br />So, the solution to the equation $3^{2x-2}=27$ is $x = \frac{5}{2}$.<br /><br />8) To solve the equation $\frac {1^{x}}{8}=4^{x+7}$, we can rewrite 1 as $1^1$ and 8 as $2^3$, and then use the property of exponents that states $a^m = a^n$ implies $m = n$.<br /><br />$\frac {1^{x}}{8}=4^{x+7}$<br /><br />$\frac {1^{x}}{2^3}=(2^2)^{x+7}$<br /><br />$\frac {1^{x}}{2^3}=2^{2(x+7)}$<br /><br />$1^{x} = 2^{3+2(x+7)}$<br /><br />$1^{x} = 2^{3+2x+14}$<br /><br />$1^{x} = 2^{2x+17}$<br /><br />Since $1^x$ is always equal to 1, we have $2^{2x+17} = 1$, which implies $2x+17 = 0$, so $x = -\frac{17}{2}$.<br /><br />So, the solution to the equation $\frac {1^{x}}{8}=4^{x+7}$ is $x = -\frac{17}{2}$.