Question 473602

{{{1+(4/3)log((x-3),4)=(11/3)}}}	    1. Add -1 to both sides of the equation.
{{{(4/3)log((x-3),4)=(11/3)-1}}}	    2. Simplify 11/3 – 1.
{{{(4/3)log((x-3),4)=(11-3)/3}}}	 
{{{(4/3)log((x-3),4)=8/3}}}		             3. Multiply both sides of the equation by ¾.
{{{(4/3)log((x-3),4)=(8/3))(3/4)}}}
{{{log((x-3),4)=(8/4)}}}		
{{{log((x-3),4)=2}}}                      4. Convert logarithm to exponential form. y=log[b]x is equivalent to x=b^y.
{{{(x-3)^2=4}}}                          5. Expand (x-3)^2. (x-3)^2=x^2-6x+9.
{{{x^2-6x+9=4}}}                      6. Add -4 to both sides.
{{{x^2-6x+9-4=4-4}}}
{{{x^2-6x+5=0}}}                      7. The resulting equation is a quadratic equation. Solve this by factoring.
{{{(x-5)(x-1)=0}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;8. Use zero product property to solve for the value(s) of x. Equate each factor to 0.</br>
x-5=0  or  x-1=0</br>
x=5    or  x=1 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;9. The values of x are 5 or 1.
x=1 should be disregarded or is not a solution because it will yield to a negative base which is 1-3=-2. Logarithm can not have a negative base.
</br>
(ANSWER: x=5)