Question

solve for t. \\(\log_{7}(7t + 12) - \log_{7}t = \log_{7}8\\) \\(t = \square\\)

Explanation:

Step1: Apply log subtraction rule

Using the property \(\log_b a - \log_b c=\log_b\frac{a}{c}\), we rewrite the left - hand side of the equation \(\log_{7}(7t + 12)-\log_{7}t=\log_{7}8\) as \(\log_{7}\frac{7t + 12}{t}=\log_{7}8\).

Step2: Use one - to - one property of logs

If \(\log_b x=\log_b y\), then \(x = y\) (for \(b>0,b
eq1,x>0,y>0\)). So we can set \(\frac{7t + 12}{t}=8\).

Step3: Solve the resulting equation

Multiply both sides of the equation \(\frac{7t + 12}{t}=8\) by \(t\) (assuming \(t
eq0\)) to get \(7t+12 = 8t\).
Subtract \(7t\) from both sides: \(12=8t - 7t\), which simplifies to \(t = 12\).
We need to check the domain: For \(\log_{7}(7t + 12)\) and \(\log_{7}t\) to be defined, \(7t+12>0\) and \(t > 0\). When \(t = 12\), \(7\times12 + 12=84 + 12 = 96>0\) and \(t=12>0\), so \(t = 12\) is in the domain.

Answer:

\(12\)