What is the answer to log_{3}(x)log_{9}(x)=1 ?

The answer to log_{3}(x)log_{9}(x)=1 is x=9^{sqrt(1/2)},x= 1/(9^{\frac{1){sqrt(2)}}}

Solve for x: log_{3}(x)log_{9}(x)=1

The solution is x=9^{sqrt(1/2)},x= 1/(9^{\frac{1){sqrt(2)}}}

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lo g_{3} (x) lo g_{9} (x) = 1

x = 9^{\frac{1}{2}}, x = \frac{1}{9 ^{\frac{1}{2}}}

Decimal Notation

x = 4.72880 \dots, x = 0.21146 \dots

Solution steps

lo g_{3} (x) lo g_{9} (x) = 1

Apply log rules

2 lo g_{9} (x)^{2} = 1

Rewrite the equation with

lo g_{9} (x) = u

2 u^{2} = 1

Solve

2 u^{2} = 1 : u = \frac{1}{2}, u = - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = lo g_{9} (x),

solve for

x

Solve

lo g_{9} (x) = \frac{1}{2} : x = 9^{\frac{1}{2}}

Solve

lo g_{9} (x) = - \frac{1}{2} : x = \frac{1}{9 ^{\frac{1}{2}}}

x = 9^{\frac{1}{2}}, x = \frac{1}{9 ^{\frac{1}{2}}}

Verify Solutions:

x = 9^{\frac{1}{2}}

True

, x = \frac{1}{9 ^{\frac{1}{2}}}

True

The solutions are

x = 9^{\frac{1}{2}}, x = \frac{1}{9 ^{\frac{1}{2}}}

ln (x) + 2 ln (2) = 2

lo g_{6} (x) + lo g_{6} (x - 16) = 2

ln (x^{2} + x) = 0

lo g_{56} (x^{2} - x) = 1

2 lo g_{3} (x) - lo g_{3} (x + 4) = 2

What is the answer to log_{3}(x)log_{9}(x)=1 ?
The answer to log_{3}(x)log_{9}(x)=1 is x=9^{sqrt(1/2)},x= 1/(9^{\frac{1){sqrt(2)}}}
Solve for x: log_{3}(x)log_{9}(x)=1
The solution is x=9^{sqrt(1/2)},x= 1/(9^{\frac{1){sqrt(2)}}}