Originally Posted by

**Shapeshift** Here are a couple of questions:

First:

You are given the equation y=logbase 10 (kx-4) + c

The x intercept is (7,0) and the graph passes through the points (52,1).

Determine the value of k and c.

** $\displaystyle y = \log_{10}{kx-4}+C $. The x intercept is (7,0). Hence:**

$\displaystyle 0 = \log_{10}{7k-4}+C $

$\displaystyle \log_{10}{7k-4}=-C $

$\displaystyle 7k-4=10^{-C} $

The other point gives:

$\displaystyle 1 = \log_{10}{52k-4}+C $

$\displaystyle \log_{10}{52k-4}=1-C $

$\displaystyle \log_{10}{52k-4}=1-C $

$\displaystyle 52k-4=10^{1-C} $

You now have two equations and two unknowns (C and k), can you solve them?

Second:

Solve for A in terms of B and C if:

A, B, and C > 1, and

logbase C (AB) - 1/logbase A (C) = logbase C (A) + logbase root12 (12)

Any help is appreciated.

$\displaystyle \log_{C}{|AB|} - \frac{1}{\log_{A}{|C|} = log_{C}{|A|} + log_{\sqrt{12}}{|12|}$