# Variable with exponential equation help

• Jul 4th 2012, 08:18 PM
Tren301
Variable with exponential equation help
Hi there, I seem to be having trouble with this equation. (X^3+5)/4=8

My idea was to multiply the values inside the bracket by 4 to undo the division, and then to do the same to the RHS...ie. 4x8, and work from there...but this gave me x= 1.4422495

But this method does not work out. I used my casio Classpad 330 to solve the equation, and x=3

Can someone give me a heads up where I am going wrong(Speechless)
• Jul 4th 2012, 08:52 PM
Soroban
Re: Variable with exponential equation help
Hello, Tren301!

Quote:

Hi there, I seem to be having trouble with this equation: . $\frac{x^3+5}{4}\:=\:8$

My idea was to multiply the values inside the bracket by 4 to undo the division . What?
and then to do the same to the RHS ... ie, 4x8, and work from there ... but this gave me x= 1.4422495

But this method does not work out. I used my casio Classpad 330 to solve the equation, and x=3

Can someone give me a heads up where I am going wrong?

We are given: . $\frac{x^3+5}{4} \:=\:8$

Multiply by 4: . ${\color{red}4}\cdot\frac{x^3+5}{4} \:=\:{\color{red}4}\cdot8$

And we have: . $x^3 + 5 \:=\:32 \quad\Rightarrow\quad x^3 \:=\:27 \quad\Rightarrow\quad x \:=\:3$
• Jul 4th 2012, 08:55 PM
pickslides
Re: Variable with exponential equation help
$\displaystyle \frac{x^3+5}{4}=8$

$\displaystyle x^3+5=8\times 4$

$\displaystyle x^3+5=32$

$\displaystyle x^3=32-5$

$\displaystyle x^3=27$

$\displaystyle x=\sqrt[3]{27} = 3$