How do you solve log_2 x + log_2 x-15 = 4?

1 Answer
Mar 31, 2016

x~~724.08

Explanation:

1. Start by adding 15 to both sides of the equation.

log_2x+log_2x-15=4

log_2x+log_2x=19

2. Use the logarithmic property, log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n) to simplify the left side of the equation.

log_2(x*x)=19

log_2(x^2)=19

3. Use the logarithmic property, log_color(purple)b(color(purple)b^color(orange)x)=color(orange)x, to rewrite the right side of the equation.

log_2(x^2)=log_2(2^19)

4. Since the equation now follows a "log=log" situation, where the bases are the same on both sides, rewrite the equation without the "log" portion.

x^2=2^19

5. Solve for x.

sqrt(x^2)=sqrt(2^(19)

x=2^(19/2)

color(green)(|bar(ul(color(white)(a/a)x~~724.08color(white)(a/a)|)))

Note that although when you take the square root of 2^19, a negative solution should also be produced, it is not actually a solution because if you substitute x=-2^(19/2) back into the original equation, you will notice that you end up taking the logarithm of a negative number, which is not possible.