# Exponential and Logarithmic Equations

• Dec 13th 2009, 01:34 PM
cheezeandbunnies
Exponential and Logarithmic Equations
how do i solve

(e^x + e^-x)/ (e^x - e^-x)

ln(log(x)) = 0

sqrt(ln x) = ln sqrt(x)

(log3X)^2 - log3X^2 = 3

ln x^2 = (lnx)^2

5^(2x) - 3(5^x) +2 = 0

ln x ^(lnx) = 4
• Dec 13th 2009, 11:17 PM
bigwave
(e^x + e^-x)/ (e^x - e^-x)
$\displaystyle \frac{(e^x + e^{-x})}{ (e^x - e^{-x})}$

this one is just an expression did you mean to set to zero

otherwise all one can is just come up with an alternate form
• Dec 13th 2009, 11:51 PM
red_dog
1) $\displaystyle \sqrt{\ln x}=\ln\sqrt{x}$

$\displaystyle \ln x\geq 0\Rightarrow x\geq 1$

The equation can be written as $\displaystyle \sqrt{\ln x}=\frac{1}{2}\ln x$

Square both members: $\displaystyle \ln x=\frac{1}{4}\ln^2x$

$\displaystyle \ln x\left(\frac{1}{4}\ln x-1\right)=0$

Now continue.

2) $\displaystyle \log_3^2x-\log_3x^2=3\Rightarrow \log_3^2x-2\log_3x-3=0$

Let $\displaystyle \log_3x=t$ and solve the quadratic.

3) $\displaystyle \ln x^2=\ln^2x\Rightarrow\ln^2x-2\ln x=0\Rightarrow\ln x(\ln x-2)=0$

4) $\displaystyle 5^{2x}-3\cdot 5^x+2=0$

Let $\displaystyle 5^x=t$ and solve the quadratic.

5) $\displaystyle \ln x^{\ln x}=4\Rightarrow\ln^2x=4\Rightarrow \ln x=\pm 2$
• Dec 14th 2009, 05:41 AM
Soroban
Hello, cheezeandbunnies!

Here's some help . . .

Quote:

$\displaystyle 1)\;\;\frac{e^x + e^{-x}}{e^x - e^{-x}}$ . . . . not an equation

Quote:

$\displaystyle 2)\;\;\ln(\log x) \:=\: 0$

We have: .$\displaystyle \ln(\log x) \:=\:0 \quad\Rightarrow\quad \log x \:=\:e^0 \:=\:1$

Therefore: .$\displaystyle \log x \:=\:1 \quad\Rightarrow\quad x \:=\:10^1 \:=\:\boxed{10}$

Quote:

$\displaystyle 3)\;\;\sqrt{\ln x} \:=\: \ln(\sqrt{x})$

We have: .$\displaystyle \sqrt{\ln x} \:=\:\ln\left(x^{\frac{1}{2}}\right) \quad\Rightarrow\quad \sqrt{\ln x} \:=\:\tfrac{1}{2}\ln x \quad\Rightarrow\quad 2\sqrt{\ln x} \:=\:\ln x$

Square both sides: .$\displaystyle 4\ln x \:=\:(\ln x)^2 \quad\Rightarrow\quad (\ln x)^2 - 4\ln x \:=\:0$

Factor: .$\displaystyle \ln x(\ln x - 4) \:=\:0$

Therefore: .$\displaystyle \begin{array}{ccccccc}\ln x \:=\:0 & \Rightarrow & \boxed{x \:=\:1} \\ \ln x - 4 \:=\:0 & \Rightarrow & \ln x \:=\:4 & \Rightarrow & \boxed{x \:=\:e^4} \end{array}$

Quote:

$\displaystyle 4)\;\;(\log_3x)^2 - \log_3x^2 \:=\: 3$

We have: .$\displaystyle (\log_3x)^2 - 2\log_3x -3 \:=\:0$

Factor: .$\displaystyle (\log_3x + 1)(\log_3x - 3) \:=\:0$

Therefore: .$\displaystyle \begin{array}{ccccccc}\log_3x+1\:=\:0 & \Rightarrow & \log_3x \:=\:-1 & \Leftarrow & \text{no real solution} \\ \\[-4mm] \log_3x-3\:=\:0 & \Rightarrow & \log_3x\:=\:3 & \Rightarrow & x \:=\:3^3 \:=\:\boxed{27} \end{array}$