I keep getting the wrong answer. Can someone please correct my working out
- a^x=b^(1-x)
- In(a)^x=In(b)^(1-x)
- xIn(a)=(1-x)In(b)
- xIn(a)=In(e)-xIn(b)
- xIn(a)+xIn(b)=In(e)
- x[In(a)+In(b)]=Ine
- x=In(e)/[(In(a)+In(b)]
the correct answer is In(b)/(In(a)+In(b)
2026-04-03 05:35:31.1775194531
Prove this logarithm equation
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How have you find $\displaystyle \ln e?$
Applying logarithm wrt $e$ on $\displaystyle a^x=b^{1-x}$
we get $$ x\ln a=(1-x)\ln b\implies x(\ln a+\ln b)=\ln b$$