# Thread: more problems I dont get.

1. ## more problems I dont get.

more problems I can't understand

9^x+2 = 27^-3

I get that 3 will go into both of them but am lost after that, notes don't help

instructions are to solve each exponential equation. Round to 2 decimal places.

and another logarithm

Log (xy^2/Z^4)
Z

edit: Z should be right after/underneath "log" and before the (.

2. Originally Posted by kenny87
more problems I can't understand

9^x+2 = 27^-3
$9^{x+2} = 27^{-3}$

$(3^2)^{x+2} = (3^3)^{-3}$

$3^{2(x+2)} = 3^{3\times -3}$

$2(x+2) = 3\times -3$

Can you finish this?

3. I guess the problem says 9^(X+2)=27^-3. In this case, (3^2)^(X+2)=(3^3)^-3 and get 3^(2X+4)=3^-9 and from that point you can go on.

4. Originally Posted by kenny87
and another logarithm

Log (xy^2/Z^4)
Z

edit: Z should be right after/underneath "log" and before the (.
What are you supposed to do with this expression?

$
\log_z \frac{xy^2}{z^4} = \log_z xy^2- \log_z z^4 = \log_z xy^2- 4\log_z z = \log_z xy^2- 4\times 1= \log_z xy^2- 4
$

5. says to expand or condense expression as much as possible.

6. more problems I can't understand

9^x+2 = 27^-3

my bad, problem is 9^x+2 = 27^ -x

I must have been looking two places at once or something. I figured that one out though looking at what you did.

It was in the first section of my notes, one more though

e^5x =77

I have one example dealing with E but it has E on both sides so not sure what to do here. Hopefully this is my last question, thanks for the help.

7. Originally Posted by kenny87

, one more though

e^5x =77
$e^{5x} =77$

$5x =\ln 77$

$x =\frac{\ln 77}{5}$

where $\ln = \log_e$

8. thanks, so I guess I would take e * 77 and divide by 5 to get my decimal answer?

9. Originally Posted by kenny87
thanks, so I guess I would take e * 77 and divide by 5 to get my decimal answer?

No, you would find $\frac{\log_e77}{5}$ to get a decimal approximation.