I have these questions on exponentials and I wanted to know if I am correct..
1. $2^{3-x} = 565$
$3-x\log 2 =\log 565$
$3-x =\frac{\log 565}{\log 2}$
$x = 3-\frac{\log 565}{\log 2}$
2. $(1+\frac{.1}{12})^{12t}=2$
$12t \log (1+12) = \log 2$
$12 t (\log 12) = \log 2$
$t = \frac{\log 2}{12 \log 12}$
3. $6e^{1-x}-25=0$
$6(1-x)\ln e = \ln 25$
$6(1-x) = \ln (5^2)$
$6(1-x) = 2\ln 5$
$x = 1-\frac{2\ln 5}{6}$
4. $e^{0.09t}-3=0$
$0.09t = \ln 3$
$t = \frac{\ln 3}{0.09}$
5. $3e^{\frac{3x}{2}} -962=0$
$e^{\frac{3x}{2}} = \frac{962}{3}$
$\frac{3x}{2} = \ln\frac{962}{3}$
$ x = 2/3 (\ln\frac{962}{3})$
2nd one seems wrong.
$(1+\frac{1}{12})^{12t} = 2$
$12t\log (1+\frac{1}{12}) = \log2$
$12t\log (\frac{13}{12}) = 12t[\log13 - \log12] = \log2$
$t = \frac{\log2}{12(\log13-\log12)}$
And the 3rd one.
$6e^{1-x} = 25$
$\ln(6e^{1-x}) = \ln(25)$
$\ln(6) + (1-x) = \ln(25)$
$\ln(6) - \ln(25) + 1 = x$
$x = \ln(\frac{6}{25}) + 1$