Solve 9^2x = 66 x=_______? Round answer to the nearest ten-thousandth place
please help
9^2 is the same thing as \[9\times9\] So it's 81x=66. Now you need to solve for x by dividing 66 by 81. 66/81 = 0.8148148...... Now you need to round to the ten-thousandths place, which would be the second 8. So x = 0.8148
incorrect but thanks for trying
i think i need to do something with log or ln
Incorrect, hmm sorry then.
Take a look at this: http://openstudy.com/study#/updates/5116c19ae4b0e554778bc9dd
\[ 9^2x = 66\]\[81x = 66\]\[x=66/81~=~22/27\]OR\[9^{\Large\ 2x~} =~ 66\]\[\log~(~9^{\Large\ 2x~} ~)=\log~(~ 66)\]\[2x~\log~(~9 ~)=\log~(~ 66)\]\[2x~=\frac{\log66}{\log8}\]\[x~(\log~10^2)=\frac{\log66}{\log8}\]\[x~(\log~100)=\frac{\log66}{\log8}\]\[x=\frac{\log~66}{\log8 \times \log 100}\]then use the calclutor to count each log, and then count entire thing. There are many ways to d this problem.
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