# Math Help - Logarithm

1. ## Logarithm

Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

Question 1
Prove that $\log_{y} x = \frac{1}{log_{x} y}$
Given that x and y are real positive numbers.

Find the possible values of x if $2(\log_ {9} x + \log_{x} 9) = 5$

Question 2
If $a^2 + b^2 = 7ab$
show that $2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b$

2. Originally Posted by ecogreen
Hi everyone... need some logarithms questions.... Thanks a lot for your help in advance. I appreciate it greatly.

Question 1
Prove that $\log_{y} x = \frac{1}{log_{x} y}$
Given that x and y are real positive numbers.

Find the possible values of x if $2(\log_ {9} x + \log_{x} 9) = 5$

Question 2
If $a^2 + b^2 = 7ab$
show that $2\log_{10} \frac{a + b}{3} = \log_{10} a + log_{10} b$

1. Using the change of base rule...

$\log_y{x} = \frac{\ln{x}}{\ln{y}}$

$= \frac{1}{\frac{\ln{y}}{\ln{x}}}$

$=\log_x{y}$.

Using this information:

$2(\log_ {9} x + \log_{x} 9) = 5$

$\log_9{x} + \log_x{9} = \frac{5}{2}$

$\log_9{x} + \frac{1}{\log_9{x}} = \frac{5}{2}$

$X + \frac{1}{X} = \frac{5}{2}$, where $X = \log_9{x}$

$X^2 + 1 = \frac{5}{2}X$

$X^2 - \frac{5}{2}X + 1 = 0$

$2X^2 - 5X + 2 = 0$

$2X^2 - 4X - X + 2 = 0$

$2X(X - 2) - 1(X - 2) = 0$

$(X - 2)(2X - 1) = 0$

$X - 2 = 0$ or $2X - 1 = 0$

$X = -2$ or $X = \frac{1}{2}$.

So that means $\log_9{x} = -2$ or $\log_9{x} = \frac{1}{2}$.

$x = 9^{-2}$ or $x = 9^{\frac{1}{2}}$

$x = \frac{1}{81}$ or $x = 3$.