# How do you solve 3^(4x) = 3^(5-x)?

May 2, 2016

$x = 1$

#### Explanation:

$\textcolor{b l u e}{\text{Quickest way using shortcuts}}$
$\textcolor{b r o w n}{\text{If you have "x^a=x^b" then } a = b}$

$\implies 4 x = 5 - x$

$4 x + x = 5$

$5 x = 5 \implies x = 1$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Another method that demonstrates other properties of indices.}}$

Note that 3^(5-x)" is the same as "3^5/(3^x

Write as ${3}^{4 x} = {3}^{5} / {3}^{x}$

Multiply both sides by ${3}^{x}$

${3}^{4 x} \times {3}^{x} = {3}^{5} \times {3}^{x} / {3}^{x}$

But ${3}^{x} / {3}^{x} = 1 \text{ and } {3}^{4 x} \times {3}^{x} = {3}^{5 x}$ giving

${3}^{5 x} = {3}^{5}$

Comparing just the indices

$5 x = 5$

$\implies x = 1$

May 2, 2016

Notice from the outset that the two exponents must be equal, since their bases are both $3$.

${\textcolor{red}{3}}^{\textcolor{b l u e}{4 x}} = {\textcolor{red}{3}}^{\textcolor{b l u e}{5 - x}} \text{ "=>" "color(blue)(4x)=color(blue)(5-x)" "=>" } 5 x = 5$

Thus, $x = 1$.

May 2, 2016

$x = 1$

#### Explanation:

As a Real-valued function of Real numbers, $f \left(x\right) = {3}^{x}$ is strictly monotonically increasing and is therefore one-to-one.

So $f \left(4 x\right) = {3}^{4 x} = {3}^{5 - x} = f \left(5 - x\right)$ implies $4 x = 5 - x$

Add $x$ to both sides to get:

$5 x = 5$

Divide both sides by $5$ to get:

$x = 1$