Question

24 (unit 4) (aks 2e) what are the solutions to the following logarithmic equations? (a) \\(\log_{4} 64 = 2x - 1\\) a \\(x=-1\\) b \\(x=0\\) c \\(x=1\\) d \\(x=2\\)

Explanation:

Step1: Evaluate the logarithm

Recall that \(\log_{a}b = c\) means \(a^{c}=b\). For \(\log_{4}64\), we know that \(4^{3}=64\), so \(\log_{4}64 = 3\).

Step2: Solve the linear equation

Substitute \(\log_{4}64 = 3\) into the equation \(\log_{4}64=2x - 1\), we get \(3=2x - 1\).
Add 1 to both sides: \(3 + 1=2x-1 + 1\), which simplifies to \(4 = 2x\).
Divide both sides by 2: \(\frac{4}{2}=\frac{2x}{2}\), so \(x = 2\). Wait, no, wait, let's check again. Wait, \(4^{3}=64\), so \(\log_{4}64 = 3\). Then the equation is \(3=2x - 1\). Add 1 to both sides: \(2x=3 + 1=4\), then \(x = 2\)? But wait, maybe I made a mistake. Wait, \(4^{2}=16\), \(4^{3}=64\), so \(\log_{4}64 = 3\). Then \(3=2x - 1\), so \(2x=4\), \(x = 2\). But let's check the options. Option D is \(x = 2\). Wait, but let's re - check the calculation.
Wait, maybe I miscalculated the logarithm. Wait, \(4^{3}=64\), so \(\log_{4}64 = 3\). Then the equation is \(3=2x - 1\). Solving for \(x\):
\(2x=3 + 1=4\)
\(x=\frac{4}{2}=2\). So the solution is \(x = 2\), which is option D.

Answer:

D. \(x = 2\)