# Exponential and Logarithmic Functions

• Dec 12th 2010, 04:22 PM
Auburn
Exponential and Logarithmic Functions
Kinda lost on this problem.

http://i.imgur.com/QRpk5.gif

Would appreciate some help. Does this have to do with log properties?

-Auburn
• Dec 12th 2010, 04:33 PM
Prove It
No need for logarithms, just convert everything to the same base.

$\displaystyle \displaystyle \frac{25^{3x^2}}{5^{7x}} = \frac{1}{25}$

$\displaystyle \displaystyle \frac{(5^2)^{3x^2}}{5^{7x}} = \frac{1}{5^2}$

$\displaystyle \displaystyle \frac{5^{6x^2}}{5^{7x}} = 5^{-2}$

$\displaystyle \displaystyle 5^{6x^2 - 7x} = 5^{-2}$

$\displaystyle \displaystyle 6x^2 - 7x = -2$

$\displaystyle \displaystyle 6x^2 - 7x + 2 = 0$

$\displaystyle \displaystyle 6x^2 - 3x - 4x + 2 = 0$

$\displaystyle \displaystyle 3x(2x - 1) - 2(2x - 1) = 0$

$\displaystyle \displaystyle (2x-1)(3x-2) = 0$

$\displaystyle \displaystyle 2x-1 = 0$ or $\displaystyle \displaystyle 3x-2 = 0$

$\displaystyle \displaystyle x = \frac{1}{2}$ or $\displaystyle \displaystyle x = \frac{2}{3}$.
• Dec 12th 2010, 04:42 PM
Auburn
Right, I see now. Since it's the same base the exponents must be equal. Thank you kindly!