# Solving an exponential equation.

• Jan 15th 2012, 01:40 PM
mido22
Solving an exponential equation.
$\displaystyle [(2x^1^/^3)-(5^1^/^3)]^3 = 35$

find X ???
• Jan 15th 2012, 02:05 PM
e^(i*pi)
Re: i want help with finding solution set of exponential equation
Quote:

Originally Posted by mido22
$\displaystyle [(2x^1^/^3)-(5^1^/^3)]^3 = 35$

find X ???

You can treat this like a normal equation to isolate 2x^(1/3):
$\displaystyle 2x^{1/3} = \sqrt[3]{35} + 5^{1/3}$
• Jan 15th 2012, 02:16 PM
mido22
Re: i want help with finding solution set of exponential equation
sorry but this is not the right solution in my book....
here is the main problem :
$\displaystyle (2x^1^/^3 - 5^1^/^3) (4x^2^/^3 + 2(5x)^1^/^3 + 25^1^/^3) = 35$

then i tried to simplify it as i wrote in the 1st post

the solution set is {5}
• Jan 15th 2012, 03:05 PM
e^(i*pi)
Re: i want help with finding solution set of exponential equation
Quote:

Originally Posted by mido22
sorry but this is not the right solution in my book....

Always state the problem in full, we can only answer what's put in front of us

Quote:

here is the main problem :
$\displaystyle (2x^1^/^3 - 5^1^/^3) (4x^2^/^3 + 2(5x)^1^/^3 + 25^1^/^3) = 35$

then i tried to simplify it as i wrote in the 1st post

the solution set is {5}
Start by expanding the brackets:

Multiplying $\displaystyle 2x^{1/3}$ by each term in the second bracket:

$\displaystyle 8x + 4\sqrt[3]{5}x^{2/3} + 2\sqrt[3]{25}x^{1/3}$

Multiplying $\displaystyle -5^{1/3}$ by each term in the second bracket

$\displaystyle -4\sqrt[3]{5}x^{2/3} - 2\sqrt[3]{25}x^{1/3} - 5$

The expanded form is the sum of these:

$\displaystyle \left(8x + 4\sqrt[3]{5}x^{2/3} + 2\sqrt[3]{25}x^{1/3}\right)$

$\displaystyle + \left(-4\sqrt[3]{5}x^{2/3} - 2\sqrt[3]{25}x^{1/3} - 5\right) = 35$

Can you go from there? I expect a few terms will cancel
• Jan 16th 2012, 12:38 AM
mido22
Re: i want help with finding solution set of exponential equation
Quote:

Originally Posted by e^(i*pi)
Always state the problem in full, we can only answer what's put in front of us

Start by expanding the brackets:

Multiplying $\displaystyle 2x^{1/3}$ by each term in the second bracket:

$\displaystyle 8x + 4\sqrt[3]{5}x^{2/3} + 2\sqrt[3]{25}x^{1/3}$

Multiplying $\displaystyle -5^{1/3}$ by each term in the second bracket

$\displaystyle -4\sqrt[3]{5}x^{2/3} - 2\sqrt[3]{25}x^{1/3} - 5$

The expanded form is the sum of these:

$\displaystyle \left(8x + 4\sqrt[3]{5}x^{2/3} + 2\sqrt[3]{25}x^{1/3}\right)$

$\displaystyle + \left(-4\sqrt[3]{5}x^{2/3} - 2\sqrt[3]{25}x^{1/3} - 5\right) = 35$

Can you go from there? I expect a few terms will cancel

thank you vm! ....................................