I know I have to take the log of both sides, but I don't seem to get how they work out!

Can anyone explain how to do

3e^(1+x) = 2

AND

10^(2x+3) = 280

If you walk me through one I should be able to figure out the other. Hopefully...

I think you need to go back to one of the log rules:

log[base A](A^n)=n

For example:

log[base10](10^3)=3

similarly

log[base e](e^-x)=-x

Does this help?

Honestly... no X_X. Don't you love college professors who don't /actually/ teach. Thanks for trying. I'm going to try and find someone who can show me.

I'm sorry if my previous explanation wasn't clear enough. I'll try to break it down step by step for you.

Let's start with the equation 3e^(1+x) = 2. The goal is to isolate x.

Step 1: Divide both sides by 3 to get rid of the coefficient in front of e^(1+x):
(3e^(1+x)) / 3 = 2 / 3

Simplifying, we have:
e^(1+x) = 2/3

Step 2: Take the natural logarithm (ln) of both sides:
ln(e^(1+x)) = ln(2/3)

Using the logarithm rule you mentioned, ln(e^(1+x)) simplifies to (1+x):
1 + x = ln(2/3)

Step 3: Subtract 1 from both sides to isolate x:
x = ln(2/3) - 1

Now let's move on to the second equation, 10^(2x+3) = 280.

Step 1: Take the logarithm base 10 (log) of both sides:
log(10^(2x+3)) = log(280)

Using the logarithm rule log(A^B) = B * log(A), we have:
(2x+3) * log(10) = log(280)

Since log(10) = 1, we can simplify to:
2x + 3 = log(280)

Step 2: Subtract 3 from both sides to isolate 2x:
2x = log(280) - 3

Step 3: Divide both sides by 2 to solve for x:
x = (log(280) - 3) / 2

Remember to use a calculator or numerical methods to evaluate the logarithms if needed.

I hope this step-by-step explanation helps you understand how to solve these equations. If you have any further questions, please let me know!