Find x x

Algebra Level 3

Solve for x . x.

3 log ⁡ 3 x 2 27 = 2 log ⁡ 3 x 9 \large 3\log_{3x^2}27 = 2\log_{3x}{9}

If your answer is a b \dfrac{a}{b} ,enter your answer as a b . ab.

Note that a , b a,b are co prime integers.


The answer is 243.

Sai Ram by Sai Ram , 19, India

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2 solutions

Sai Ram
Oct 28, 2015

The given equation is 3 log ⁡ 3 x 2 27 = 2 log ⁡ 3 x 9 3\log_{3x^2}27 = 2\log_{3x}9

This can be written as

3 log ⁡ 3 x 2 27 = 2 log ⁡ 3 x 9 ⇒ log ⁡ 3 x 2 ( 3 3 ) 3 = log ⁡ 3 x ( 3 2 ) 2 \large 3\log_{3x^2}27 = 2\log_{3x}9 \Rightarrow \log_{3x^2}(3^3)^3 = \log_{3x}(3^2)^2

⇒ log ⁡ 3 x 2 3 9 = log ⁡ 3 x 3 4 ⇒ 9 4 = log ⁡ 3 x 3 log ⁡ 3 x 2 3 ⇒ 9 4 = log ⁡ 3 x 3 x 2 . \large \Rightarrow \log_{3x^2}3^9 = \log_{3x}3^4 \Rightarrow \dfrac{9}{4} = \dfrac{\log_{3x}3}{\log_{3x^2}3} \Rightarrow \dfrac{9}{4} = \log_{3x}{3x^2}.

Now,

( 3 x ) 9 4 = 3 x 2 . \large (3x)^\dfrac{9}{4} = 3x^2.

Simplifying it further, we get,

( 3 ) 5 4 × x 1 4 = 1 ⇒ ( 3 5 ) 1 4 × ( x ) 1 4 = 1 ⇒ x = 1 243 \large (3)^{\dfrac{5}{4}} \times x^{\dfrac{1}{4}}= 1 \Rightarrow (3^5)^{\dfrac{1}{4}} \times (x)^{\dfrac{1}{4}} = 1 \Rightarrow \boxed{\boxed{x=\dfrac{1}{243}}}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Pranay Kumar
Sep 11, 2015

=> log base(3x^2) 3^9 = log base(3x) 3^4

=> 9/4 = log base(3x^2) 3 / log base(3x) 3

=> 9/4 = log base(3x) 3x^2

=> (3x)^9/4 = 3x^2

=> 3^(5/4) * x^(1/4) =1

=> x = 1/243 :) :)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You can use latex.

Sai Ram - 5 years, 9 months ago

Your solution is wrong in the second step.

Once check my solution.

Sai Ram - 5 years, 7 months ago

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