SOLUTION: Solve for x using logs 10^(x+3)=5E^(7-x) 9^x=2e^x2 pls help

Question 205210: Solve for x using logs 10^(x+3)=5E^(7-x)

9^x=2e^x2

pls help
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Solve for x using logs
10^(x+3) = 5 * e^(7-x)
;
using the natural logs
ln(10^(x+3)) = ln(5) + ln(e^(7-x)}
:
log equiv of exponents
(x+3)ln(10) = ln(5) + (7-x)ln(e^}
:
find the natural logs, (nat log of e is 1)
2.30(x+3) = 1.61 + 7 - x
:
2.3x + 6.9 = 8.61 - x
:
2.3x + x = 8.61 - 6.9
:
3.3x = 1.7
x =
x = .5
:
Check on a calc x=.5:
10^3.5 = 3262
5*e^6.5 = 3326
Does not check out exactly on a calc because I dropped all those decimals
you can use the method & retain the decimals and get a more exact value for x
:
:
9^x = 2 * e^(x^2)
do this the same way
x*ln(9) = ln(2) + x^2*ln(e)
2.20x = .69 + x^2
:
0 = x^2 - 2.2x + .69
Use the quadratic formula to find x: a=1; b=-2.2 c=.69
two solutions x~.38 and x~1.8
:

RELATED QUESTIONS
answer using logs:... (answered by Alan3354)

Solve for x using logs.... (answered by nerdybill)

Solve for x.... (answered by lwsshak3)

help!!!!... (answered by Tatiana_Stebko)

solve for x using logs... (answered by Tatiana_Stebko)

Solve for x: 12/1+5e^(-x) =... (answered by Fombitz)

Solve e^x = 3 - 2e^-x I have a problem regarding natural logs since they're a little (answered by solver91311)

Solve for X 10^(x+3)= 5e^(7-x) I have only gotton to here: 2^(x+3)=... (answered by Nate)

solve for x... (answered by Fombitz,nyc_function)