Question

log(2x+8)=1+log(x-3) AND
log(x)+log(x-3)=1

I am not sure how to do these.

Answer 1

\[\log_{10?}x+\log_{10}y= \log_{10}xy \]\[\log_{10}x-\log_{10y}=\log_{10}(x/y)\]

\[\log_{10}b=x \]so \[b=10^{x}\]

So we should be able to figure this out using these logarithm properties.

So for the first problem we have \[\log_{10}(2x+8)=1+\log_{10}(x-3) \] So subtract from both sides. \[\log_{10}(x-3) \] This gives us\[\log_{10}(2x+8)-\log_{10}(x-3)=1 \]

Using the first property we have\[\log_{10}[(2x+8)/(x-3)]=1 \] So \[(2x+8)(x-3)=10^{1}\]

Now solve for x using the quadratic equation.

Try doing the second one on your own. If you get stuck, I'll be around for another hour or someone else can jump in.