Question 397589: log(6*5^x-25*20^x)=x+log(25)
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! log(6*5^x-25*20^x)=x+log(25)
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log(6*5^x-25*20^x)-log(25)=x
=log{(6*5^x-25*20^x)/25}=x
change to exponential form
10^x={(6*5^x-25*20^x)/25}
(2^x*5^x)=((6*5^x)-25(4^x*5^x))/25
divide by 5^x
2^x=(6-25*4^x)/25
25*2^x=6-25*4^x
25*4^x+25*2^x=6
25*4^x+25*2^x-6=0
4^x=2^2x
let u=2^x
then u^2=2^2x
you then have a quadratic equation to solve
25u^2+25u-6=0
(5u+6)(5u-1)=0
5u+6=0
u=-6/5 (reject, no solution}
5u-1=0
u=1/5
u=2^x
2^x=1/5=.2
xlog2=log(.2)
x=(log(.2))/log2
from calculator
ans: x=-2.322
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Check:
x+log25=-2.322+1.398=-.924
6*5^x=6*5^-2.322=.143
25*20^x=25*20^-2.322=.024
log(.143-.024)=log(.119)=-.924
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