I am trying to solve these questions:
1. $8^x=4$
$\log 8^4 = x$ and this becomes $4\log 8$
$\log 8 = \log 2^3 = 3\log 2$ so it finally becomes $12 \log 2$
2. $4e^{2x} = 91$
$(e^{x})^{2} = \frac{91}{4}$
$e^x = \frac{\sqrt{91}}{2}$
$x = \ln \frac{\sqrt{91}}{2}$
3. $\log (x+4)-\log x = \log (x+2)$
$\log\frac{x+4}{x} = \log(x+2)$
$10^{\log\frac{x+4}{x}} = 10^{x+2}$
$\frac{x+4}{x} = x+2$
$x = \frac{-1\pm \sqrt{17}}{2}$
4. $\log_4 x - \log_4 (x-1) = \frac{1}{2}$
$4^{\log_4 \frac{x}{x-1}} = 4^{1/2}$
$\frac{x}{x-1} = 2$
$x = 2$
5. $\frac{3000}{e^{2x} +2} = 2$
$3000= 2e^{2x}+4$
$1498 = e^{2x}$
$\ln 1498 = 2x $
$x = \frac{\ln1498}{2}$
Are these answers correct?
Also can someone please show me how to graph these problems? I know that there are a lot but maybe just how to graph number 5 or number 4. Or just provide me with a hint so i can graph them myself.
Thank you!