# need help with 2 Logarithm problems

• Dec 2nd 2008, 05:46 AM
Sonx
need help with 2 Logarithm problems
hey i got a test today and i dont know how to do these 2 problems:

$\displaystyle 5^(2x) - 6(5)^x + 5 = 0$

AWNSER: x = 1 or 0

$\displaystyle 2log(base4)(x+1) - log(base4)(2x-3) = 2$

AWNSER: x = 28.3 or 1.7
• Dec 2nd 2008, 06:47 AM
Soroban
Hello, Sonx!

Quote:

$\displaystyle 5^{2x} - 6\!\cdot\!5^x + 5 \:=\: 0$
That ugly expression is a quadratic.

It factors! . . $\displaystyle (5^x - 1)(5^x-5) \:=\:0$

So we have: .$\displaystyle \begin{array}{ccccc}5^x-1\:=\:0 & \Rightarrow & 5^x\:=\:1 &\Rightarrow& \boxed{x \:=\:0} \\ \\[-3mm] 5^x-5\:=\:0 &\Rightarrow&5^x\:=\:5 &\Rightarrow &\boxed{ x \:=\:1} \end{array}$

Quote:

$\displaystyle 2\log_4(x+1) - \log_4(2x-3) \:= \:2$

We have: .$\displaystyle \log_4(x+1)^2 - \log_4(2x-3) \:=\:2 \quad\Rightarrow\quad \log_4\left[\frac{(x+1)^2}{2x-3}\right] \;=\;2$

. . . $\displaystyle \frac{(x+1)^2}{2x-3} \:=\:4^2 \quad\Rightarrow\quad (x+1)^2 \:=\:16(2x-3)$

. . . $\displaystyle x^2+2x+1 \:=\:32x - 48 \quad\Rightarrow\quad x^2-30x + 49 \:=\:0$

Quadratic Formula: . $\displaystyle x \;=\;\frac{\text{-}(\text{-}30) \pm\sqrt{(\text{-}30)^2 - 4(1)(49)}}{2(1)} \;=\;\frac{30 \pm\sqrt{704}}{2} \;=\;15 \pm4\sqrt{11}$

Answers: .$\displaystyle x \;=\;\begin{Bmatrix}15 + 4\sqrt{11} &=& 28.26649916 \\ \\[-4mm]15 - 4\sqrt{11} &=& 1.733500839 \end{Bmatrix}$