Question

4^(log 2x base 2) = 36
Find x

Answer 1

Simplifiying LHS,
${4}^{ log_{2}(2x) }$
$= {2}^{2 log_{2}(2x) }$
$= {2}^{log_{2}(4x^{2})} \\ = 4x^{2}$
Hence, 4x²= 36
Which implies,
X=±3

Answer 2

Answer:  The required value of x is 3.

Step-by-step explanation:  We are given to find the value of x from the following logarithmic equation :

$4^{\log_22x}=36~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following properties of logarithms :

$(i)~a^{\log_ax}=x,\\\\(ii)~a\log b=\log b^a.$

From equation (i), we have

$4^{\log_22x}=36\\\\\Rightarrow (2^2)^{\log_22x}=36\\\\\Rightarrow 2^{2\log_22x}=36\\\\\Rightarrow 2^{\log_2(2x)^2}=36\\\\\Rightarrow (2x)^2=36\\\\\Rightarrow 4x^2=36\\\\\Rightarrow x^2=\dfrac{36}{4}\\\\\Rightarrow x^2=9\\\\\Rightarrow x=\pm\sqrt{9}~~~~~~~~~~~~~~~[\textup{taking square root on both sides}]\\\\\Rightarrow x=\pm3.$

Since logarithm of a negative number does not exist, so we get

x = 3.

Thus, the required value of x is 3.