Question 196795
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{{{9^x=4}}}

Take logs of both side:

{{{log(9^x)=log(4)}}}

Use a rule of logarithms on the left

{{{x*log(9)=log(4)}}}

Divide both sides by {{{log(9)}}}

{{{(x*log(9))/(log(9))=(log(4))/(log(9))}}}

{{{(x*cross(log(9)))/(cross(log(9)))=(log(4))/(log(9))}}}

{{{x = .6020599913/.9542425094}}}

{{{x = .6309297536}}}

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{{{9^x = 3/3^x}}}

Write {{{9}}} as {{{3^2}}}

Write {{{3}}} on the right as {{{3^1}}}

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

Multiply inner exponent by outer exponent on
the left side.

Subtract the exponents on the right:

{{{3^(2x)=3^(1-x)}}}

Since the bases on each side are positive and
not equal to 1, we can equate the exponents:

{{{2x=1-x}}}

{{{2x+x=1}}}

{{{3x=1}}}

{{{x=1/3}}}

Edwin</pre>