Urgent log functions

**leviathanwave** · January 30th 2007, 02:01 PM

Hi, i mixed up my log properties a bit and got these problems wrong.

1. find the inverse of the function y = logbase7(x + 3) + 7 <--- for this one i know that i have to interchange the x and y, but i dont know where to go from there

2. Solve the following for x and leave in exact form. Check for extraneous roots.

a. ln(2x - 2) - ln(x-1) = ln(x)

b. (log(8x) - 2log(2x))/logx = 1

c. 2ln(3x) + 5 = 8

d. log(x) = 90 - log(x - 90)

e. x^2(log(x)) - 4(log(x)) = 0 <--- for this one there is a hint, it says to factor first, but i can't seem to figure it out.

Any help is much appreciated, thanks a lot.

**Soroban** · January 30th 2007, 09:19 PM

Hello, leviathanwave!

1. Find the inverse of: . $y \:= \:\log_7(x + 3) + 7$

"Switch" $x$ and $y\!:\;\;x \;=\;\log_7(y + 3) + 7$

Solve for $y\!:\;\;x - 7 \;=\;\log_7(y + 3)$

. . . . . . . . . $y + 3 \;=\;7^{x-7}$

. . . . . . . . . . . . $y \;=\;7^{x-7} - 3$

2. Solve the following for x and leave in exact form. Check for extraneous roots.

Note that $\log(x)$ exists for $x > 0$
. . For other values, the log does not exist (DNE).

$a)\;\;\ln(2x - 2) - \ln(x-1) \:= \:\ln(x)$

We have: . $\ln\left(\frac{2x-2}{x-1}\right)\:=\:\ln(x)$

"Un-log": . $\frac{2x-2}{x-1} \:=\:x\quad\Rightarrow\quad 2x - 2 \;=\;x^2 - x$

We have a quadratic: . $x^2 - 3x + 2 \;=\;0$

. . which factors: . $(x - 1)(x - 2)\;=\;0$

. . and has roots: . $x \:=\:1,\,2$

For $x = 1$ , DNE.

The only root is: . $\boxed{x = 2}$

$b)\;\;\frac{\log(8x) - 2\log(2x)}{\log(x)} \:= \:1$

We have: . $\log(8x) - \log(2x)^2\;=\;\log(x)\quad\Rightarrow\quad \log(8x) - \log(4x^2) \;=\;\log(x)$

. . $\log\left(\frac{8x}{4x^2}\right)\;=\;\log(x)\quad\ Rightarrow\quad \log\left(\frac{2}{x}\right)\;=\;\log(x)$

Un-log: . $\frac{2}{x}\;=\;x\quad\Rightarrow\quad x^2 \,= \,2\quad\Rightarrow\quad x \,=\,\pm\sqrt{2}$

For $x = -\sqrt{2}$ , DNE.

The only root is: . $\boxed{x\,=\,\sqrt{2}}$

$c)\;\;2\ln(3x) + 5 \:= \:8$

We have: . $\ln(3x)^2\;=\;4$

Then: . $(3x)^2\;=\;e^3\quad\Rightarrow\quad 3x\;=\;\pm e^{\frac{3}{2}}\quad\Rightarrow\quad x\:=\:\pm\frac{1}{3}e^{\frac{3}{2}}$

For $x = -\frac{1}{3}e^{\frac{3}{2}}$ , DNE.

The only root is: . $\boxed{x\,=\,\frac{1}{3}e^{\frac{3}{2}}}$

$d)\;\;\log(x) \:= \:90 - \log(x - 90)$

Is there a typo?. We can't solve this one . . .

We have: . $\log(x) + \log(x-90)\;=\;90\quad\Rightarrow\quad \log\left[x(x - 90)\right]\;=\;90$

Then: . $x(x - 90)\;=\;10^{90}\quad\Rightarrow\quad x^2 - 90x - 10^{90} \;=\;0$ ??

$e)\;\;x^2\log(x) - 4\log(x) \: = \:0$

Factor: . $\log(x)\!\cdot\!(x^2 - 4)\;=\;0$

Factor: . $\log(x)\!\cdot\!(x - 2)(x + 2)\;=\;0$

Then: . $\begin{Bmatrix}\log(x) = 0& \Rightarrow & x = 1 \\ x-2 \:=\:0 & \Rightarrow & x = 2 \\ x + 2\:=\:0 & \Rightarrow & x = \text{-}2\end{Bmatrix}$

For $x = \text{-}2$ , DNE.

The roots are: . $\boxed{x \:=\:1,\,2}$

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