Question

log base 9 ( 3 log base 2( 1 + log base 3 (1+ 2 log x base 2)))=1/2

Answer 1

Solution!!

$\sf \log _{9}(3\log _{2}(1+\log _{3}(1+2\log _{2}(x))))=\dfrac{1}{2}$

Convert the logarithm into exponential form using the fact that $\sf \log _{a}(x)=b\:is\:equal\:to\:x=a^{b}$.

$\sf 3\log _{2}(1+\log _{3}(1+2\log _{2}(x)))=9^{\frac{1}{2}}$

Write the number in exponential form with the base of 3.

$\sf 3\log _{2}(1+\log _{3}(1+2\log _{2}(x)))=(3^{2})^{\frac{1}{2}}$

Simplify the expression by multiplying exponents.

$\sf 3\log _{2}(1+\log _{3}(1+2\log _{2}(x)))=3$

Divide both sides of the equation by 3.

$\sf \log _{2}(1+\log _{3}(1+2\log _{2}(x)))=1$

Convert the logarithm into exponential form again using the fact that $\sf \log _{a}(x)=b\:is\:equal\:to\:x=a^{b}$.

$\sf 1+\log _{3}(1+2\log _{2}(x))=2^{1}$

Any expression raised to the power of 1 equals itself.

$\sf 1+\log _{3}(1+2\log _{2}(x))=2$

Move the constant to the right-hand side and change its sign.

$\sf \log_{3}(1+2\log _{2}(x))=2-1$

Subtract the numbers.

$\sf \log_{3}(1+2\log _{2}(x))=1$

Convert the logarithm into exponential form again using the fact that $\sf \log _{a}(x)=b\:is\:equal\:to\:x=a^{b}$.

$\sf 1+2\log _{2}(x)=3^{1}$

Any expression raised to the power of 1 equals itself.

$\sf 1+2\log _{2}(x)=3$

Move the constant to the right-hand and change its sign.

$\sf 2\log _{2}(x)=3-1$

Subtract the numbers.

$\sf 2\log _{2}(x)=2$

Divide both sides of the equation by 2.

$\sf \log _{2}(x)=1$

Convert the logarithm into exponential form again using the fact that $\sf \log _{a}(x)=b\:is\:equal\:to\:x=a^{b}$.

$\sf x=2^{1}$

Any expression raised to the power of 1 equals itself.

$\sf x=2$