Question

Can anyone help me solve these three equations?
_=base

log_2(x+1)+log_2(x-2)=2

AAND

x^4-2x^3=4x^2-8x

AAND

(2/x)-([x+3]/[x+5])=(-2/[x^2+5x])

Answer 1

\[\log_2(x+1)+\log_2(x-2)=2 \]
use \(\log(A) +\log(B)=\log(AB) \) to write as
\[\log_2((x+1)(x-2))=2\] then rewrite in exponential form as
\[(x+1)(x-2)=2^2=4\]

Answer 2

solve the quadratic equation, then make sure to check your answers because it is likely that one will not work

Answer 3

Okay, I got that one. Thanks :). I actually just got the second one on my list on my own, but can you help me with the third?

Answer 4

The third one is\[\frac{2}{x}-\frac{x+3}{x+5}=\frac{-2}{x^{2}+5x}\]?

First, you want to combine the two fractions on the left side. This can be done by multiplying the first fraction (2/x) by the denominator of the second fraction over itself, and then multiplying the second fraction by the denominator of the first fraction over itself, and then adding them together. So\[\frac{2}{x}\times \frac{x+5}{x+5}-\frac{x+3}{x+5}\times \frac{x}{x}=\frac{2x+10}{x^{2}+5x}-\frac{x^{2}+3x}{x^{2}+5x}=\frac{-x^{2}-x+10}{x^{2}+5x}\]Then we have\[\frac{-x^{2}-x+10}{x^{2}+5x}=\frac{-2}{x^{2}+5x}\]Notice that the denominators are now equal to each other. This allows you to solve the much easier equation\[-x^{2}-x+10=-2\]with the following restriction: x cannot be such that \[x^{2}+5x=0\]

Answer 5

Thanks!