Solving equations with exponents (1 Viewer)

kels15 · Oct 20, 2019

Hi guys, just asking for some help with a question. How do I solve: 2^x-3/4^1-x = 1? Thanks!

Drdusk · Oct 20, 2019

Is it

$\dfrac{3}{4^{1-x}}$

or

$\bigg(\dfrac{3}{4}\bigg)^{1-x}$

?

kels15 · Oct 20, 2019

Drdusk said:
Is it

$\dfrac{3}{4^{1-x}}$

or

$\bigg(\dfrac{3}{4}\bigg)^{1-x}$

?

It's

Screen Shot 2019-10-20 at 6.47.02 pm.png

Drdusk · Oct 20, 2019

Come on guys at least format your question properly. It's not that hard.

$\dfrac{2^{x-3}}{4^{1-x}} = 1$

$\therefore \dfrac{2^{x-3}}{2^{2(1-x)}} = 1$

$\therefore 2^{3x-5} = 1$

$(3x-5)ln(2) = 0 \Rightarrow 3x-5 = 0\Rightarrow x = \dfrac{5}{3}$

vernburn · Oct 20, 2019

Solution without logs (just in case you don’t know them!):
$\begin{aligned}\frac{2^{x-3}}{4^{1-x}}&=1\\ \Rightarrow 2^{x-3}&=4^{1-x}\\ 2^{x-3}&=2^{2-2x}\\ x-3&=2-2x \\ 3x&=5 \\ \therefore x&=\frac53\end{aligned}$

kels15 · Oct 20, 2019

vernburn said:
Solution without logs (just in case you don’t know them!):
$\begin{aligned}\frac{2^{x-3}}{4^{1-x}}&=1\\ \Rightarrow 2^{x-3}&=4^{1-x}\\ 2^{x-3}&=2^{2-2x}\\ x-3&=2-2x \\ 3x&=5 \\ \therefore x&=\frac53\end{aligned}$

This is exactly what I needed! Thank you!

akkjen · Oct 21, 2019

Drdusk said:
Come on guys at least format your question properly. It's not that hard.

$\dfrac{2^{x-3}}{4^{1-x}} = 1$

$\therefore \dfrac{2^{x-3}}{2^{2(1-x)}} = 1$

$\therefore 2^{3x-5} = 1$

$(3x-5)ln(2) = 0 \Rightarrow 3x-5 = 0\Rightarrow x = \dfrac{5}{3}$

vernburn said:
Solution without logs (just in case you don’t know them!):
$\begin{aligned}\frac{2^{x-3}}{4^{1-x}}&=1\\ \Rightarrow 2^{x-3}&=4^{1-x}\\ 2^{x-3}&=2^{2-2x}\\ x-3&=2-2x \\ 3x&=5 \\ \therefore x&=\frac53\end{aligned}$

how do you guys type maths symbols like tht?

fan96 · Oct 21, 2019

akkjen said:
how do you guys type maths symbols like tht?

You need to learn the TeX typesetting system: https://en.wikipedia.org/wiki/TeX.

In the forums you can use tex tags to properly typeset math symbols, e.g.

Code:

[tex]I = \int_0^\infty e^{-x^2} \, dx.[/tex]

gives this:

$I = \int_0^\infty e^{-x^2} \, dx.$

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Solving equations with exponents (1 Viewer)

kels15

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