Question

@jdoe0001 How do you solve for the equation log x+ log(x-3) = 1?

Answer 1

@jdoe0001 I know the answer is x= .5(3+sqrt of (9+4e)). I just don't know how the answer key got that answer.

Answer 2

hmmm

Answer 3

@jdoe0001 actually I just graphed it on my graphing calculator. x=5

Answer 4

right... I get something else
one sec

Answer 5

ok @jdoe0001

Answer 6

\(log(x)+log(x-3)=1\implies log[(x)(x-3)]=1
\\ \quad \\
log_{{\color{brown}{ 10}}}[(x)(x-3)]=1\implies {\color{brown}{ 10}}^1=(x)(x-3)
\\ \quad \\
10=x^2-3x\implies 0=x^2-3x-10
\\ \quad \\
0=(x-5)(x+2)\implies
\begin{cases}
x=5\\
x=-2
\end{cases}\)

Answer 7

@jdoe0001 are you sure about x=-2? I plugged -2 in for x and I got "Math ERROR" in the calculator

Answer 8

right... logarithms do not take a negative value, is 0 or up
so ends up being 5 :)

Answer 9

but what the answer sheet shows.. looks very different