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

find X ???

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- Jan 15th 2012, 01:40 PMmido22Solving an exponential equation.
$\displaystyle [(2x^1^/^3)-(5^1^/^3)]^3 = 35$

find X ??? - Jan 15th 2012, 02:05 PMe^(i*pi)Re: i want help with finding solution set of exponential equation
- Jan 15th 2012, 02:16 PMmido22Re: 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 PMe^(i*pi)Re: i want help with finding solution set of exponential equation
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}

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 AMmido22Re: i want help with finding solution set of exponential equation