# How do you solve 4^x - 2^x = 0?

Apr 2, 2018

$x = 0$

#### Explanation:

${4}^{x} - {2}^{x} = 0$

${4}^{x} = {2}^{x}$

Apply the natural logarithm to both sides:

$\ln \left({4}^{x}\right) = \ln \left({2}^{x}\right)$

Recalling that $\ln \left({a}^{b}\right) = b \ln \left(a\right)$:

$x \ln \left(4\right) = x \ln \left(2\right)$

So, $x = 0$ is obviously a solution as it will result in $0 = 0.$

Beyond that, there are no solutions as $x$ cancels out and we're left with

$\ln \left(4\right) = \ln \left(2\right)$ which is not true.

Apr 2, 2018

$x = 0$

#### Explanation:

${4}^{x} - {2}^{x} = 0$

now
$4 = {2}^{2} \implies {4}^{x} = {2}^{2 x}$

${4}^{x} - {2}^{x} = 0 \implies {2}^{2 x} - {2}^{x} = 0$

factorising

${2}^{x} \left({2}^{x} - 1\right) = 0$

either  2^x=0=>no real solns

or

${2}^{x} - 1 = 0$

${2}^{x} = 1 \implies x = 0$

Apr 2, 2018

The answer is $x = 0$.

#### Explanation:

The other answers on the page are correct; I just wanted to display another method to solve this problem:

${4}^{x} - {2}^{x} = 0$

${4}^{x} = {2}^{x}$

${\left({2}^{2}\right)}^{x} = {2}^{x}$

${2}^{\textcolor{b l u e}{2 x}} = {2}^{\textcolor{b l u e}{x}}$

Since the bases are equivalent, the exponents must also be equivalent:

$\textcolor{b l u e}{2 x} = \textcolor{b l u e}{x}$

$x = 0$

Apr 3, 2018

Real solution: $x = 0$

Complex solutions: $x = \frac{2 k \pi i}{\ln} 2 \text{ }$ for any integer $k$.

#### Explanation:

Given:

${4}^{x} - {2}^{x} = 0$

Note that ${4}^{x} = {\left(2 \cdot 2\right)}^{x} = {2}^{x} \cdot {2}^{x}$

So we have:

$0 = {4}^{x} - {2}^{x} = {2}^{x} \left({2}^{x} - 1\right)$

Note that ${2}^{x} = 0$ has no solutions (real or complex).

So:

${2}^{x} = 1$

This has real solution $x = 0$

Note that ${e}^{2 \pi i} = 1$, so ${e}^{2 k \pi i} = 1$ for any integer $k$. These are the only complex values for which ${e}^{t} = 1$.

So we find:

${e}^{2 k \pi i} = 1 = {2}^{x} = {\left({e}^{\ln 2}\right)}^{x} = {e}^{x \ln 2}$

So:

$x \ln 2 = 2 k \pi i$

So:

$x = \frac{2 k \pi i}{\ln} 2 \text{ }$ for any integer $k$