Could someone please solve these questions for me? Thanks
Hello, strangepath!
That's quite a list, but I see that they require a variety of tecnniques.
I'll solve a few of them.
$\displaystyle \text{(1) Solve for }n\!:\;\;11^{n+1} \:=\:13$
Take logs: .$\displaystyle \ln(11^{n+1}) \:=\:\ln(13) \quad\Rightarrow\quad(n+1)\ln(11) \:=\:\ln(13)$
. . . . . . . . .$\displaystyle n+1 \:=\:\frac{\ln(13)}{\ln(11)} \quad\Rightarrow\quad n \:=\:\frac{\ln(13)}{\ln(11)} - 1$
$\displaystyle \text{(4) Solve for }x\!:\;\;5^{x+1} - 5^x \:=\:8$
This one is deceptively simple.
Factor: .$\displaystyle 5^x(5-1) \:=\:8 \quad\Rightarrow\quad 5^x\cdot4 \:=\:8 \quad\Rightarrow\quad 5^x \:=\:2$
Take logs: .$\displaystyle \ln(5^x) \:=\:\ln(2) \quad\Rightarrow\quad x\ln(5) \:=\:\ln(2) \quad\Rightarrow\quad x \:=\:\frac{\ln(2)}{\ln(5)}$
$\displaystyle \text{(7) Solve for }x\!:\;\;10^{4\log x} \:=\:16$
We are expected to know that: .$\displaystyle 10^{4\log x} \:=\:10^{\log x^4} \:=\:x^4$
The equation becomes: .$\displaystyle x^4 \:=\:16 \quad\Rightarrow\quad x \:=\:2$
. . . We must discard x = -2.
$\displaystyle \text{(8) Solve for }w\!:\;\;3e^w - 35e^{-w} \:=\:-16$
Multiply by $\displaystyle e^w\!:\;\;3e^{2w} - 35 \:=\:-16e^w \quad\Rightarrow\quad 3e^{2w} + 16e^w - 35 \:=\:0$
Factor: .$\displaystyle (e^w + 7)(3e^w - 5) \:=\:0$
And we have two equations to solve:
. . $\displaystyle \begin{Bmatrix}e^w + 7 \:=\:0 & \Rightarrow & e^w \:=\:\text{-}7 &\Rightarrow & \text{no real roots} \\ \\ 3e^w - 5 \:=\:0 & \Rightarrow& e^w \:=\:\frac{5}{3} & \Rightarrow & w \:=\:\ln(\frac{5}{3}) \end{Bmatrix}$