kim solved the equation below by graphing a system of equations. log₂(3x - 1) = log₄(x + 8) what is the…

Explanation:

Step1: Change the base of the logarithm

We know that (\log_{a^n}b=\frac{1}{n}\log_ab). So, (\log_{4}(x + 8)=\log_{2^2}(x + 8)=\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)) Using the property (n\log_{a}M=\log_{a}M^{n}), we can rewrite it as (\log_{2}(3x - 1)=\log_{2}\sqrt{x + 8}) Since if (\log_{a}m=\log_{a}n), then (m = n) (for (a>0,a\neq1,m>0,n>0)), we have (3x-1=\sqrt{x + 8})

Step2: Square both sides

Let (y=\sqrt{x + 8}), then (y^{2}=x + 8) and (x=y^{2}-8). The equation (3x-1=\sqrt{x + 8}) becomes (3(y^{2}-8)-1=y) (3y^{2}-y-25 = 0) Using the quadratic formula (y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ay^{2}+by + c=0) (here (a = 3), (b=-1), (c=-25)) (y=\frac{1\pm\sqrt{1+300}}{6}=\frac{1\pm\sqrt{301}}{6}) Since (y=\sqrt{x + 8}\geq0), (y=\frac{1+\sqrt{301}}{6}\approx\frac{1 + 17.35}{6}\approx3.06) If (y=\sqrt{x+8}\approx3.06), then (x=y^{2}-8\approx9.36-8 = 1.36\approx1.4)

Another way: Let (y_1=\log_{2}(3x-1)) and (y_2=\log_{4}(x + 8)) When (x = 1.4) (y_1=\log_{2}(3\times1.4-1)=\log_{2}(4.2-1)=\log_{2}(3.2)\approx1.7) (y_2=\log_{4}(1.4 + 8)=\log_{4}(9.4)=\frac{\log_{2}(9.4)}{\log_{2}(4)}=\frac{\log_{2}(9.4)}{2}\approx\frac{3.2}{2}=1.6\approx1.7)

Answer:

(1.4)