Question

kim solved the equation below by graphing a system of equations.\\(\log_{2}(3x - 1)=\log_{4}(x + 8)\\)\
what is the approximate solution to the equation?\
\\(\bigcirc\\) 0.6\
\\(\bigcirc\\) 0.9\
\\(\bigcirc\\) 1.4\
\\(\bigcirc\\) 1.6

Explanation:

Step1: Rewrite the equations

We can rewrite the equation \(\log_{2}(3x - 1)=\log_{4}(x + 8)\) as a system of equations: \(y = \log_{2}(3x - 1)\) and \(y=\log_{4}(x + 8)\). We can also use the change - of - base formula or logarithmic properties to solve it algebraically and then check with the given options.

First, recall that \(\log_{a^n}b=\frac{1}{n}\log_{a}b\). So \(\log_{4}(x + 8)=\log_{2^2}(x + 8)=\frac{1}{2}\log_{2}(x + 8)=\log_{2}\sqrt{x + 8}\)

The original equation \(\log_{2}(3x - 1)=\log_{2}\sqrt{x + 8}\) (since if \(\log_{a}m=\log_{a}n\), then \(m = n\) for \(a>0,a
eq1,m>0,n>0\))

So we have the equation \(3x-1=\sqrt{x + 8}\)

Step2: Test the given options

• For \(x = 0.6\):

Left - hand side of \(3x - 1\): \(3\times0.6-1=1.8 - 1=0.8\)
Right - hand side: \(\sqrt{0.6 + 8}=\sqrt{8.6}\approx2.93\)
Since \(0.8
eq2.93\), \(x = 0.6\) is not a solution.

• For \(x = 0.9\):

Left - hand side: \(3\times0.9-1 = 2.7-1=1.7\)
Right - hand side: \(\sqrt{0.9 + 8}=\sqrt{8.9}\approx2.98\)
Since \(1.7
eq2.98\), \(x = 0.9\) is not a solution.

• For \(x = 1.4\):

Left - hand side: \(3\times1.4-1=4.2 - 1 = 3.2\)
Right - hand side: \(\sqrt{1.4+8}=\sqrt{9.4}\approx3.07\)
The two sides are relatively close.

• For \(x = 1.6\):

Left - hand side: \(3\times1.6-1=4.8 - 1=3.8\)
Right - hand side: \(\sqrt{1.6 + 8}=\sqrt{9.6}\approx3.1\)
Since \(3.8
eq3.1\)

We can also solve the equation \(3x - 1=\sqrt{x + 8}\) by squaring both sides (note that we have to check for extraneous solutions later):

\((3x - 1)^2=x + 8\)

\(9x^{2}-6x + 1=x + 8\)

\(9x^{2}-7x - 7 = 0\)

Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for the quadratic equation \(ax^{2}+bx + c = 0\), here \(a = 9\), \(b=-7\), \(c=-7\)

\(x=\frac{7\pm\sqrt{(-7)^{2}-4\times9\times(-7)}}{2\times9}=\frac{7\pm\sqrt{49 + 252}}{18}=\frac{7\pm\sqrt{301}}{18}\)

\(\sqrt{301}\approx17.35\)

\(x=\frac{7 + 17.35}{18}\approx\frac{24.35}{18}\approx1.35\approx1.4\) (we discard the negative root since \(3x-1>0\) and \(x + 8>0\) for the domain of the logarithmic functions)

Answer:

1.4 (the option corresponding to 1.4, e.g., if the options are A. 0.6, B. 0.9, C. 1.4, D. 1.6, then the answer is C. 1.4)