How do i solve these exponential equations?

Question

Is there a way to solve these exponential equations without using logarithms?
I tried to get the same base for all the terms, but I could not make it.

Is there any other general procedure that I can use to solve them?

Thank you for your help.

$(e^x - \frac{1}{e})(e^{2x}-\sqrt{e}) = 0 $

$ 8^{x^2-2x} = \frac{1}{2}$

$10^x + 10^{x+1} = \frac{11}{10}$

Answer 1

For your first question

Hint

You are look for values of $x$ such that the factors: $$(e^x - \frac{1}{e})(e^{2x}-\sqrt{e}) = 0$$ That is: $$e^x = e^{-1}, \ e^{2x} = \sqrt{e}$$

Answer 2

For the first one,

$$(e^x-e^{-1})(e^{2x}-e^{0.5})=0$$

$$\implies e^x-e^{-1}=0,e^{2x}-e^{0.5}=0$$

$$\left\{ \begin{aligned} e^x&=e^{-1} \\ e^{2x}&=e^{0.5} \\ \end{aligned} \right.$$

For the second one,

$$8^{x^2-2x}=\frac12$$

$$2^{2(x^2-2x)}=2^{-1}$$

For the last one,

$$10^x+10^{x+1}=\frac{11}{10}$$

$$10^x+10\cdot10^x=\dots$$

$$(1+10)10^x=\dots$$

$$11\cdot10^x=\frac{11}{10}$$

$$10^x=\frac1{10}$$

$$10^x=10^{-1}$$

Answer 3

I'm not sure what you think logorithms are that need avoiding but :

1)$(e^x - \frac{1}{e})(e^{2x}-\sqrt{e})= (e^x - e^{-1})(e^{2x} - e^{1/2} = 0$

so $e^x - e^{-1} = 0$ or $e^{2x} - e^{1/2} = 0$

So $x = -1$ or $x=1/4$

2) $8^{x^2 - 2x} = 1/2 = 2^{-1} = 8^{-1/3}$

3) $10^x + 10^{x+1} = 10^x(1 + 10) = 10^x*11 = 11/10$ so $10^x = 1/10 = 10^{-1}$.