Question

What is the true solution to the logarithmic equation below?
log₂ (6x)-log₂ (√x)=2

Answer 1

answer: x =  $\\ \frac{4}{9\\}$

$log_{2}$ (6x) - $log_{2}$ ($\sqrt{x}$) = 2

expand the expression:

• $log_{2}$(6x) = $log_{2}$ (6) + $log_{2}$ (x)

• $log_{2}$ ($\sqrt{x}$) = $log_{2}$ ($x^{\frac{1}{2} }$)

transform the expression:

• $log_{2}$ ($x^{\frac{1}{2} }$) = $\frac{1}{2}$ * $log_{2}$ (x)

calculate the difference:

• $log_{2}$ (x) - $\frac{1}{2}$ * $log_{2}$ (x) = $\frac{1}{2}$ * $log_{2}$ (x)

multiply both sides by 2:

• $log_{2}$ (6) + $\frac{1}{2}$ * $log_{2}$ (x) = 2   is now   2 $log_{2}$ (6) + $log_{2}$ (x) = 4

transform the expression:

• 2 $log_{2}$ (6) = $log_{2}$ ($6^{2}$)

simplify the expression:

• $log_{2}$ ($6^{2}$) + $log_{2}$ (x) = $log_{2}$ ($6^{2} x$)

evaluate the power:

• $log_{2}$ ($6^{2} x$) = $log_{2} (36x)$

convert the logarithm into exponential form:

• $log_{2}(36x) = 4$    is now   $36x = 2^{4}$

evaluate the power:

• $2^{4} = 16$

divide both sides by 36:

• $\frac{36}{36}$ x  =  $\frac{16}{36}$   which is   x = $\frac{4}{9}$