Question

solve the equation. log₈x + log₈(x + 30) = 2 select the correct choice below and fill in any answer boxes present a. x = (simplify your answer, including any radicals. use integers or b. there is no solution.

Explanation:

Step1: Apply log - product rule

Using the rule $\log_aM+\log_aN = \log_a(MN)$, we rewrite the left - hand side of the equation $\log_8x+\log_8(x + 30)$ as $\log_8[x(x + 30)]$. So the equation becomes $\log_8[x(x + 30)]=2$.

Step2: Convert to exponential form

By the definition of logarithms, if $\log_a b=c$, then $b=a^c$. So, $x(x + 30)=8^2$. Expanding the left - hand side gives $x^2+30x=64$. Rearranging to standard quadratic form, we get $x^2+30x - 64=0$.

Step3: Solve the quadratic equation

For a quadratic equation $ax^2+bx + c = 0$ (here $a = 1$, $b = 30$, $c=-64$), we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$. First, calculate the discriminant $\Delta=b^2-4ac=(30)^2-4\times1\times(-64)=900 + 256=1156$. Then $x=\frac{-30\pm\sqrt{1156}}{2}=\frac{-30\pm34}{2}$.
We have two solutions:
$x_1=\frac{-30 + 34}{2}=\frac{4}{2}=2$ and $x_2=\frac{-30-34}{2}=\frac{-64}{2}=-32$.

Step4: Check for valid solutions

Since the argument of a logarithm must be positive, for $y = \log_8x$, $x>0$ and for $y=\log_8(x + 30)$, $x+30>0$. When $x = 2$, $\log_82$ and $\log_8(2 + 30)=\log_832$ are well - defined. When $x=-32$, $\log_8(-32)$ is not a real number. So we discard $x=-32$.

Answer:

A. $x = 2$