1. Urgent log functions

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.

2. Hello, leviathanwave!

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

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

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

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

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

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

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

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

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

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

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

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

For $\displaystyle x = 1$, DNE.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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