Given that 2log2(x+15) -log2(x) = 6, show that x^2-34x+225=0

As we know nlog(a) is the same as log(a)^n, we can rearrange the equation to log2(x+15)^2 -log2(x) = 6. Also, we know that log(a) - log(b) is equal to log(a/b), so we can further rearrange the equation to log2((x+15)^2)/x) = 6. Another law of logarithms states to find a, when logx(a) = b , a = x^b. Thus, the equation becomes ((x+15)^2)/x) = 2^6Now we can simplify this equation. Opening the brackets of the numerator gives us x^2 +30x +225, all of which is divisible by x. 2^6 = 64. Because (x^2 +30x +225)/x = 64 we can multiply everything by x, giving us x^2 +30x +225 = 64xNow to rearrange it in the format the question asks, we can take 64x away from both sides, leaving us with x^2 -34x +225 = 0