Method to solve this type of equation

Question

It seems simple yet I have to show how I got to the answer... I've been Googling, but can't narrow it down.

$8 = 2^x$

Thanks in advance.

Answer 1

you can try plugging in values of $x=1$, $x=2$, ... and so on to see what value of x works.

algebraically, you can take $log_2(8)$ which says "what power of 2 gives me 8?"

Answer 2

This kind of equations are solved by taking a logarithms. As a logarithm base you use the number that's raised to power $x$. So in this case we have:

$$8 = 2^x$$ $$\log_2 8 = \log_2 2^x$$ $$3 = x$$

Also you can use the fact that $8 = 2^3$. So two numbers raised to some number and having same bases can be same iff their exponents are same so we have:

$$8 = 2^x \iff 2^3 = 2^x \iff x = 3$$

Answer 3

$$ 2^x = 8 $$ Taking $log_{e} $ both side.

$$ x\times log_{e}{2} = log_{e}{8}$$ $$\implies x = \frac{log_{e}{8}}{log_{e}{2}}$$ $$ \implies x = \frac{log_{e}{2^3}}{log{e}{2}}$$ $$\implies x = 3$$

Answer 4

Using the "plug in values until one works" method, but a bit more formally, define

$$a_n:=2*a_{n-1}, a_0=1$$

Then we have $a_1=2*1, a_2=2*2, a_3=2*4=8,a_4=2*8,\cdots$ Then we note that our sequence $\{a_n\}$ can also be written as $a_n=2^n$, in which case we have $a_3=2^3=8$. Further, note that $\{a_n\}$ is monotone increasing (which means that there is exactly one value of $x$ which achieves the required value $2^x=8$) thus showing that $2^x=8\iff x=3$.

Note that the answers specifying "$\log_2 8=x$" as the way to solve this problem in general are entirely correct, but they assume that the logarithm function is defined for you as the inverse-exponential function.