# Math Help - logarithm equations! need a little help solving...

1. ## logarithm equations! need a little help solving...

i'm having trouble solving a few of these logarithm problems

1) If loga2 = logb16, show that b=a^4

i'm also having trouble with squaring logs...for example one of the questions was:
(log10x)^2 - 2log10x + 1 = 0
the answer is 10, but i keep ending up canceling out both xs

the last one was(2log2x)^2 - 3log2x - 10 = 0
how exactly do you end up with two values for x in that example? the answers are x = 4, 2^(-5/4)

thanks!!

2. $\displaystyle \log_y{x}=\frac{\log{x}}{\log{y}}$

$\displaystyle \frac{\log{2}}{\log{a}}=\frac{\log{16}}{\log{b}}\R ightarrow \frac{\log{2}}{4\log{2}}=\frac{\log{a}}{\log{b}}$

$\displaystyle \frac{1}{4}=\frac{\log{a}}{\log{b}}\Rightarrow \frac{1}{4}\log{b}=\log{a}\Rightarrow 10^{\log{b^{1/4}}}10^{\log{a}}$

$\displaystyle b^{1/4}=a\Rightarrow b=a^4$

3. Do you mean this

$\displaystyle (\log_{10}{x})^2 - 2\log_{10}{x} + 1 = 0$

or this

$\displaystyle (\log{10x})^2 - 2\log{10x} + 1 = 0$

4. the first one, 10 is the base

5. $\displaystyle (\log_{10}{x})^2 - 2\log_{10}{x} + 1 = 0\Rightarrow (\log{x}-1)^2=0\Rightarrow \log{x}-1=0$

$\mbox{Let} \ (\log{x})^2=y^2 \ \ \log{x}=y\Rightarrow y^2-2y+1=0\Rightarrow (x-1)^2=0 \ \mbox{but} \ y=\log{x}$

$\Rightarrow\log{x}=1\Rightarrow 10^{log{x}}=10^1\Rightarrow x=10$

6. $(2\log_{2}{x})^2 - 3\log_{2}{x} - 10 = 0$

Is two the base?

7. yes, 2 is the base
thanks for the other answers ^_^

8. $(2\log_{2}{x})^2 - 3\log_{2}{x} - 10 = 0$

$\mbox{Let} \ y=\log_{2}{x} \ \Rightarrow 4y^2-3y-10=0\Rightarrow (y-2)(4y+5)=0$

$y=2 \ \mbox{and} \ \frac{-5}{4}$

$\displaystyle \log_2{x}=2\Rightarrow 2^2=x\Rightarrow x=4$

$\displaystyle \log_2{x}=\frac{-5}{4}\Rightarrow 2^{-5/4}=x\Rightarrow x=2^{-5/4}$

9. thanks!!

10. Your welcome. Whenever you have logs of the same base in a quadratic appearance, try to factor it.