hello, how are you.

could you please go to this link http://www.accd.edu/sac/math/faculty...i/PdfT2Rev.pdf and help me to solve those exercises, I know it sucks opening PDF document, I'd copy the exercises but I just couldn't copy them as clear as they are in that link.
thank you.

2. Hello, jhon!

You're expected to know the basic rules of logs and exponentials.

Here's the last part . . .

Solve for $x$.

$1)\;\;e^{x+1}\:=\:25$
Take logs of both sides: . $\ln\left(e^{x+1}\right) \:=\:\ln(25)$

Then: . $(x+1)\!\cdot\!\ln(e) \:=\:\ln(25)$

Since $\ln(e) = 1$, we have: . $x + 1\:=\:\ln(25)$

Therefore: . $\boxed{x \;=\;\ln(25) - 1}$

$2)\;\;5^{1-2x} \:=\:2^{x+1}$
Take logs of both sides: . $\ln\left(5^{1-2x}\right) \;=\;\ln\left(2^{x+1}\right)$

Then: . $(1-2x)\!\cdot\!\ln(5) \;=\;(x+1)\!\cdot\!\ln(2)$

Expand: . $\ln(5) - 2x\!\cdot\!\ln(5) \;=\;x\!\cdot\!\ln(2) + \ln(2)$

Rearrange: . $-2x\!\cdot\!\ln(5) - x\!\cdot\!\ln(2) \;=\;-\ln(5) + \ln(2)$

Factor: . $-x\left[2\!\cdot\!\ln(5) + \ln(2)\right] \;=\;-\left[\ln(5) - \ln(2)\right]$

Therefore: . $\boxed{x \;=\;\frac{\ln(5) - \ln(2)}{2\!\cdot\!\ln(5) + \ln(2)}}$

The last three can be solved without logs . . .

$3)\;\;5^{x+1} \:=\:625$
We have: . $5^{x+1} \:=\:5^4$

Therefore: . $x+1 \:=\:4\quad\Rightarrow\quad\boxed{x \:=\:3}$

$4)\;\;\left(\frac{1}{2}\right)^{x-1} \:=\:\frac{1}{256}$
We have: . $\frac{1}{2^{x-1}} \;=\;\frac{1}{2^8}\quad\Rightarrow\quad 2^{-(x-1)} \:=\:2^{-8}$

Therefore: . $-(x-1)\:=\:-8\quad\Rightarrow\quad x-1\:=\:8\quad\Rightarrow\quad\boxed{x = 9}$

$5)\;\;4^{2x} \:=\:32^{x-1}$
We have: . $(2^2)^{2x} \;=\;(2^5)^{x-1}\quad\Rightarrow\quad 2^{4x} \;=\;2^{5x-5}$

Therefore: . $4x \:=\:5x-5\quad\Rightarrow\quad\boxed{x \,=\,5}$

3. ## thank you

God bless you soroban, thanks a lot for your help. I'm sorry to continue bothering you, but could you please help me to solve the remaining exercises, and also if you could be so nice and explain me in easy words the basic rules of logs and exponents. Yes, I have those rules in my textbook, but is hard for me to understand them, I'm taking this class online, so I don't have a professor to ask my interrogates but I do have to take tests in the school.

Thank you for everything, please continue helping me Soroban.