Question

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

Answer 1

ANSWER

The true solution is

$x = \frac{4}{9}$
EXPLANATION

The logarithmic equation given to us is

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



We need to use the quotient rule of logarithms.

$log_{a}( M ) - log_{a}( N ) = log_{a}( \frac{ M }{ N } )$


When we apply this law the expression becomes

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


We now take the antilogarithm of both sides to get

$\frac{6x}{ \sqrt{x} } = {2}^{2}$



$\frac{6x}{ \sqrt{x} } = 4$


We square both sides to get,


$(\frac{6x}{ \sqrt{x} }) ^{2} = {4}^{2}$

We evaluate to obtain,

$\frac{36 {x}^{2} }{ x } = 16$



This simplifies to


$36x = 16$



We divide both sides by 36 to get

$x = \frac{16}{36}$
We simplify to get,
$x = \frac{4}{9}$

Answer 2

Answer:

OPTION C *x=4/9*

Step-by-step explanation: