Question

log8(x^2-30)=log8(x)

Answer 1

as there are logs both sides:
x^2 - 30 = x
solve this quadratic and you're home

Answer 2

equate the terms since they are logs of the same base

x^2 - 30 = x

x^2 - x - 30 = 0

(x - 6)(x+5) = 0

solve for x

Answer 3

Make sure x value is positive!

Answer 4

\[\LARGE \log_8(x^2-30)=\log_8(x) \quad \iff x^2-30=x\]
\[\LARGE x^2-x-30=0 \]
use quadratic formula ,
\[\LARGE x_{1/2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
or you can factor it out, if you want it's up to you :)

Answer 5

You can subtract log8(x) from both sides and get\[\log_8(x^2-30)-\log_8(x)=0\]Using the subtraction property for logs we get\[\log_8((x^2-30)/x)=0\]Going to exponential form you get \[8^0=(x^2-30)/x\]\[x=x^2-30\]\[x^2-x-30=0\]\[(x-6)(x+5)=0\]\[x=6, x=-5\] Because a logarithm is undefined for negative values of x, only x=6 is valid.

Answer 6

Thank you everyone!