Asked by Zoe
Solve exactly.
a)3^x-2=5^x
b)2^k-2=3^k+1
a)3^x-2=5^x
b)2^k-2=3^k+1
Answers
Answered by
Reiny
I will assume you meant
3^(x-2) = 5^x
or else we would have ourselves a real mess
take log of both sides and use rules of logs
(x-2)log3 = x log5
x log3 - 2log3 = x log5
x(log3 - log5) = 2log3
x = 2log3/(log3-log5)
= appr -4.30
check:
LS = 3^(-4.3 -2) = .000985
RS = 5^(-4.3) = .00985 , not bad
try the second one the same way , assuming you meant:
2^(k-2)=3^(k+1)
3^(x-2) = 5^x
or else we would have ourselves a real mess
take log of both sides and use rules of logs
(x-2)log3 = x log5
x log3 - 2log3 = x log5
x(log3 - log5) = 2log3
x = 2log3/(log3-log5)
= appr -4.30
check:
LS = 3^(-4.3 -2) = .000985
RS = 5^(-4.3) = .00985 , not bad
try the second one the same way , assuming you meant:
2^(k-2)=3^(k+1)
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