Question

How do you solve ${2^x} = 10$?

Answer 1

Hint: Here we will solve these kinds of more complicated equations, we will have to use the logarithms. Taking logarithms will allow us to take advantage of the log rule that says that powers inside a log can be moved out in front as multipliers. By taking the log of an exponential, we can then move the variable (being in the exponent that’s not inside a log) out in front, as a multiplier on the log. In other words the log rule will let us move the variable back down onto the ground, where we get our hands on it.

Complete step-by-step solution:
Given equation is ${2^x} = 10$
If this equation had asked us to solve ${2^x} = 8$ then finding the solution would have been easy. Because we could have converted the $8$ to ${2^3}$ , set the exponents equal and solved for “$x = 3$” . But unlike $8$, $10$ is not a power of $2$ so we can’t set powers equal to each other. We need some other method of getting at the $x$, because we can’t solve the equation with the variable floating up there above $2$, we need it back down on the ground where it belongs, where we can get at it. And we will have to use logarithms to bring that variable down.
When dealing with equations, we can do whatever we like to do the same thing to both sides. And to solve an equation, we have to get the variable by itself on the side of the “equals” sign, to isolate the variable, we have to “undo” whatever has been done to the variable.
In this case the variable $x$ has been put in the exponent. The backwards of exponentials are logarithms.
So multiply $\log $ on both sides of the equation, we get,
$\log \left( {{2^x}} \right) = \log \left( {10} \right)$
We know the property of logarithm that it, $\log {m^n} = n\log m$ ,
Using this property we get, \[x\log \left( 2 \right) = \log \left( {10} \right)\]
Our targeting variable is $x$ so taking other term in the left side of the equals,
$ \Rightarrow x = \dfrac{{\log \left( {10} \right)}}{{\log \left( 2 \right)}}$
We know the value of $\log 10 = 1,\,\log 2 = 0.3010$ we get,
$x = \dfrac{1}{{0.3010}} = 3.32226$

Therefore the value of $x$ is $3.32226$

Note: If $a$ is a positive real number other than $1$ and, then $x$ is called the logarithm of $m$ to the base $a,$ written as ${\log _a}m.$${\log _a}m$ exists only, if $m,a > 0{\text{ and a}} \ne {\text{1}}{\text{.}}$
If $m < 0$ then ${\log _a}$ will be imaginary.
 Logarithms were historically used and created because they make products into sums and powers into multiplications.
 So whenever you are working with an expression that involves a lot of products and powers. But not to many sums, it might be easier to take the log.