SOLUTION: Seems like this should be simple...
"Solve for x using logs. Give an exact answer.
4 * 2^x = 6 * 5^x "
I tried:
4*log2^x=6*log5^x
log2^x^4=log5^x^6
log2^4x=log5^6x
log (2
Algebra.Com
Question 154437: Seems like this should be simple...
"Solve for x using logs. Give an exact answer.
4 * 2^x = 6 * 5^x "
I tried:
4*log2^x=6*log5^x
log2^x^4=log5^x^6
log2^4x=log5^6x
log (2^4x)/(5^6x)=0
10^0=(2^4x)/(5^6x)
1=(2^4x)/(5^6x)
so x=0
Which is incorrect. I'd really appreciate some help on this one!
T
Found 2 solutions by scott8148, stanbon:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
when taking logs, exponents become coefficients
log(4) + x log(2) = log(6) + x log(5)
subtracting xlog(2)+log(6) __ log(4)-log(6)=xlog(5)-xlog(2) __ factoring __ log(2/3)=x(log(5/2))
dividing by log(5/2) __ log(2/3)/log(5/2)=x
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
4 * 2^x = 6 * 5^x
----------------
2^x/5^x = 6/4
(2/5)^x = 3/2
x*log(2/5) = log(3/2)
x = [log(3/2)]/[log(2/5)]
x = -0.442507...
Cheers,
Stan H.
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