Nov
21
Can someone explain to me how you solve this exponential function?
By
2(3^x) = 5^(x – 1)
Please help me!
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4 Comments
November 21st, 2009 at 7:10 am
Use the fact that 5^(x-1) = (5^x)/(5^1) to rewrite the equation as
2(3^x) = (5^x)/5 or
10(3^x) = 5^x
then take the log of both sides
1 + x log 3 = x log 5
the rest is straightforward.
November 21st, 2009 at 7:32 am
2(3^x) = 5^(x – 1)
2(3^x) =(5/5) 5^(x – 1)
2(3^x) =(1/5) 5^x
3^x/5^x = 1/10
(3/5)^x = 1/10
x log(3/5) = log(1/10)
x = -1/(log 3 -log 5)
x = 1/(log 5-log 3)
November 21st, 2009 at 8:07 am
Take logs
log(2) + xlog(3) = (x – 1)log)5)
x(log(3) – log(5)) = -log(5) – log(2) = -(log(5) + log(2)
xlog(3/5) = -log(10)
x = -log(10)/log(3/5)
November 21st, 2009 at 8:49 am
well u solve 3^x first…… 3×3x3=27
2(27)=54
54=5^(x-1)….thats all ive got