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Solve the exponential equation in terms of natural logarithms
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\[4^{1-x} = 5\] So I took the natural log of both sides: \[\ln (4^{1-x}) = \ln(5)\] Moved the exponent to the front: \[(1-x)\ln(4) = \ln(5)\] divided by ln(4) \[(1-x) = \frac{ \ln(5) }{ \ln(4) }\]
Where do I go from here?
So wouldn't the answer just be \[1-\frac{ \ln5 }{ \ln4 }\]
why does the one go in front?
\[1-x=\frac{ \ln5 }{ \ln4 }\] \[-x=-1+\frac{ \ln5 }{ \ln4 }\]
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So \[x=1-\frac{ \ln5 }{ \ln4 }\]
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