Question 1204568
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To type {{{log(9,(x))}}} on a keyboard, you could write log(9,x)


So something like {{{log(9,(3*sqrt(3)))}}} would be log(9, 3*sqrt(3) )



Now onto the problem at hand.
{{{log(9,(3*sqrt(3))) = x}}}


{{{9^x = 3*sqrt(3)}}} Convert the equation to exponential form


{{{(3^2)^x = 3^1*(3)^(1/2)^""}}} Rewrite 9 as 3^2. Rewrite the square root as an exponent of 1/2.


{{{3^(2x) = 3^1*(3)^(1/2)^""}}} Use the rule that (a^b)^c = a^(b*c)


{{{3^(2x) = 3^(1+1/2)^""}}} Use the rule that a^b*a^c = a^(b+c)


{{{3^(2x) = 3^(3/2)^""}}}


{{{2x = 3/2}}} Since the bases are both 3, the exponents must be equal.


{{{x = (3/2)*(1/2)}}}


{{{x = 3/4}}}


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Another approach would be to use the change of base formula.
{{{log(9,(3*sqrt(3))) = x}}}


{{{x = log(9,(3*sqrt(3)))}}}


{{{x = log((3*sqrt(3)))/(log((9)))}}} Change of base formula is applied here. From this point onward, the logs are base 10.


{{{x = log((3^1*(3)^(1/2)))/(log((3^2)))}}}


{{{x = log((3^(1+1/2)^""))/(log((3^2)))}}}


{{{x = log((3^(3/2)^""))/(log((3^2)))}}}


{{{x = ((3/2)*log((3)))/(2*log((3)))}}} Use the log rule log(A^B) = B*log(A)


{{{x = ((3/2)*highlight(log((3))))/(2*highlight(log((3))))}}}


{{{x = ((3/2)*cross(log((3))))/(2*cross(log((3))))}}}


{{{x = (3/2)/(2)}}}


{{{x = matrix(1,3,3/2,"divide",2)}}}


{{{x = (3/2)*(1/2)}}}


{{{x = 3/4}}}
Your steps do not have to be as verbose as shown above. 


As quick confirmation, use a calculator to find that:
log(3*sqrt(3))/log(9) = 0.75
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