SOLUTION: How to solve the equation using logarithms 64(1.06^2x-1)=4

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Question 863587: How to solve the equation using logarithms 64(1.06^2x-1)=4
Answer by lwsshak3(11628) About Me  (Show Source):
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How to solve the equation using logarithms
64(1.06^2x-1)=4
1.06^2x-1=4/64=1/16
1.06^2x=1/16+1=17/16
2xlog(1.06)=log(17/16)
2x=log(17/16)/log1.06)≈1.0404…
x≈0.5202…