How do you solve $16^{x - 5} = 9^{x + 1}$ $16^(x-5) = 9^(x+1)$ ?

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Answer 1 · 2016-04-21T17:31:08

$x = \frac{ln 1048576 + ln 9}{ln 16 - ln 9} \approx 27.91$ $x=(ln1048576+ln9)/(ln16-ln9)~~27.91$

${ln 16}^{x - 5} = {ln 9}^{x + 1}$ $ln 16^(x-5)=ln9^(x+1)$

$(x - 5) ln 16 = (x + 1) ln 9$ $(x-5)ln16=(x+1)ln9$

$x ln 16 - 5 ln 16 = x ln 9 + ln 9$ $xln16-5ln16=xln9+ln9$

$x ln 16 - {ln 16}^{5} = x ln 9 + ln 9$ $xln16-ln16^5=xln9+ln9$

$x ln 16 - ln 1048576 = x ln 9 + ln 9$ $xln16-ln 1048576=xln9+ln9$

$x ln 16 - x ln 9 = ln 1048576 + ln 9$ $xln16-xln9=ln1048576+ln9$

$x (ln 16 - ln 9) = ln 1048576 + ln 9$ $x(ln16-ln9)=ln1048576+ln9$

$x = \frac{ln 1048576 + ln 9}{ln 16 - ln 9} \approx 27.91$ $x=(ln1048576+ln9)/(ln16-ln9)~~27.91$

How do you solve 16x−5=9x+116^(x-5) = 9^(x+1)?