How do you solve $9^{4 x + 1} = 64$ $9^(4x+1)=64$ ?

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Answer 1 · 2018-03-09T06:35:40

$x=0.223...$

Apply log to both sides and then expand the brackets.

$(4x+1)log9=log64$
$4xlog+log9=log64$

minus $log+9$ from both sides, then divide by $log9$

$4xlog9=log64-log9$

$4x=(log64-log9)/log9$

$4x=0.8927...$
$x=0.223...$

Check the solution by subbing in the value for $x$

$9^(4(0.223...)+1$ $=64$
$64=64$

How do you solve 9^(4x+1)=6494x+1=649^(4x+1)=64?