kim solved the equation below by graphing a system of equations. $log_2(3x - 1)=log_4(x + 8)$ what is the…

Explanation:

Step1: Change base of logarithm

Use the change - of - base formula $\log_{a^n}b=\frac{1}{n}\log_ab$. So, $\log_{4}(x + 8)=\frac{\log_{2}(x + 8)}{\log_{2}4}=\frac{1}{2}\log_{2}(x + 8)$. The original equation $\log_{2}(3x-1)=\log_{4}(x + 8)$ becomes $\log_{2}(3x - 1)=\frac{1}{2}\log_{2}(x + 8)$. Then, by the power rule of logarithms $n\log_{a}M=\log_{a}M^{n}$, we have $\log_{2}(3x - 1)=\log_{2}\sqrt{x + 8}$.

Step2: Remove logarithms

Since $\log_{a}M=\log_{a}N$ implies $M = N$ (for $a>0,a\neq1,M>0,N>0$), we get $3x-1=\sqrt{x + 8}$. Square both sides to get $(3x - 1)^2=x + 8$. Expand the left - hand side: $9x^{2}-6x + 1=x + 8$. Rearrange to get a quadratic equation: $9x^{2}-7x - 7 = 0$.

Step3: Use quadratic formula

The quadratic formula for $ax^{2}+bx + c = 0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 9$, $b=-7$, and $c=-7$. So, $x=\frac{7\pm\sqrt{(-7)^{2}-4\times9\times(-7)}}{2\times9}=\frac{7\pm\sqrt{49 + 252}}{18}=\frac{7\pm\sqrt{301}}{18}\approx\frac{7\pm17.35}{18}$. We have two solutions for $x$ from the quadratic formula: $x_1=\frac{7 + 17.35}{18}\approx1.35$ and $x_2=\frac{7-17.35}{18}\approx - 0.57$. But for the original logarithmic equations $\log_{2}(3x - 1)$ and $\log_{4}(x + 8)$, we need $3x-1>0$ and $x + 8>0$. The negative solution $x\approx - 0.57$ is extraneous.

Answer:

1.4