SOLUTION: how to solve (4/3)^1-x =5^x
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Question 376396: how to solve (4/3)^1-x =5^x
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I presume you mean:
^{1\,-\,x}\ =\ 5^x)
Next time use parentheses appropriately, that is: (4/3)^(1 - x) = 5^x would have made your meaning clear.
Take the log of both sides. Whether you use natural logs or base 10 logs (or indeed any other base) doesn't matter. I use 
, just because it tickles my fancy to do so.
^{1\,-\,x}\right)\ =\ \ln\left(5^x\right))
Use
\ =\ n\log_b(x))
to write
\ln\left(\frac{4}{3}\right)\ =\ x\ln\left(5\right))
Use ordinary algebra to write:
}{\ln\left(\frac{4}{3}\right)})
Use
\ =\ \log_b(x) - \log_b(y))
To write:
}{\ln(4)\ -\ \ln(3)})
Use ordinary algebra to write:
}{\ln(4)\ -\ \ln(3)}\ =\ 1)
}{\ln(4)\ -\ \ln(3)}\right)\ =\ 1)
}{\ln(4)\ -\ \ln(3)}})
\ -\ \ln(3)}{\ln(5)\ +\ \ln(4)\ -\ \ln(3))
Use your calculator to find the numeric approximation and to check your answer.
John
My calculator said it, I believe it, that settles it
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