Alg9.4.29

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Solve $\log_2(9^{x-1}+7)=2+\log_2(3^{x-1}+1)\,$

Readjusting the logarithms

$\log_2(9^{x-1}+7)-\log_2(3^{x-1}+1)=2\,$

Applying the division rule,

$\log_2(\frac{9^{x-1}+7}{3^{x-1}+1})=2\,$

Taking out the log on lefthand side,

$\frac{3^{2x-2}+7}{3^{x-1}+1}=4\,$

$3^{2x-2}+7=4(3^{x-1}+1)\,$

$3^{2x-2}+7=4(3^{x-1}+1)\,$

Simplifying

$3(3^{2x})=36(3^x)-81\,$

Let

$3^x=k\,$

Then

$3k^2-36k+81=0\,$

Solving

$k^2-12k+27=0\,$

$(k-9)(k-3)=0\,$

$k=9,k=3\,$

Hence

$3^x=9,3^x=3\,$

The values of x are 2 and 1

Main Page:Algebra:Logarithmic

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