SOLUTION: 4e^x+2=32 ln(5x-3)^1/3=2 2x^(3x-2)+4=16 2e^(x-2)=e^x+7 1/3lnx+ln2-ln3=3 ln(x+2)-ln4=3 Hello. Thank you for your time. I have a test and Im practicing on these pro

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: 4e^x+2=32 ln(5x-3)^1/3=2 2x^(3x-2)+4=16 2e^(x-2)=e^x+7 1/3lnx+ln2-ln3=3 ln(x+2)-ln4=3 Hello. Thank you for your time. I have a test and Im practicing on these pro      Log On


   

Question 586189: 4e^x+2=32
ln(5x-3)^1/3=2
2x^(3x-2)+4=16
2e^(x-2)=e^x+7
1/3lnx+ln2-ln3=3
ln(x+2)-ln4=3
Hello. Thank you for your time. I have a test and Im practicing on these problems, but I keep getting something wrong. Can you work them out please? Thank you sooo much!!! =)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
4e^x + 2 = 32
4e^x = 32 - 2
4e^x = /4
e^x = 7.5
use nat logs here
ln(e^x) = ln(7.5)
the log equiv of exponents
x*ln(e) = ln(7.5)
ln of e is 1 therefore
x = ln(7.5}
x = 2.015
:
:
ln(5x-3)^1/3 = 2
Cube both sides
ln(5x-3) = 2^3
ln(5x-3) = 8
find the anti-log of both sides (e%5Ex)
5x - 3 = 2981
5x = 2981 + 3
x = 2984/5
x = 596.8
:
:
2x^(3x-2) + 4 = 16
2x^(3x-2) = 16 - 4
2x^(3x-2) = 12
x^(3x-2) = 12/2
x^(3x-2) = 6
use common logs
log(x^(3x-2)) = log(6)
(3x-2)log(x) = .778
log(x) = .778%2F%28%283x-2%29%29
That's about all I can do with this
:
:
2e^(x-2) = e^x + 7
2e^(x-2) - e^x = 7
I don't think there is a solution here
:
:
1%2F3ln(x) + ln(2) - ln(3) = 3
ln(x^(1/3)) + ln(2) - ln(3) = 3
combine as a single log
:
ln%28%282x%5E%281%2F3%29%29%2F3%29 = 3
find the antilog (e^10) of 3, gets rid of ln
%28%282x%5E%281%2F3%29%29%2F3%29 = 20.0855
:
2x%5E%281%2F3%29 = 3(20.0855)
:
2x%5E%281%2F3%29 = 60.2566
:
x%5E%281%2F3%29 = 60.2566%2F2
;
x%5E%281%2F3%29 = 30.1283
Cube both sides:
x = 30.1283^3
x ~ 27348