Solve this exponential equation

• Feb 5th 2009, 08:29 PM
Shandy
Solve this exponential equation
If someone could help me to solve this, I would appreciate it a lot.

(3/5)^x-1 * (2/3)^x = 40/225
• Feb 5th 2009, 08:45 PM
Jhevon
Quote:

Originally Posted by Shandy
If someone could help me to solve this, I would appreciate it a lot.

(3/5)^x-1 * (2/3)^x = 40/225

Here's a start

$\left( \frac 35 \right)^{x - 1} \cdot \left( \frac 23 \right)^x = \frac {40}{225}$

$\Rightarrow \left( \frac 35 \right)^{-1} \cdot \left( \frac 35 \right)^x \cdot \left( \frac 23 \right)^x = \frac {40}{225}$

$\Rightarrow \frac 53 \cdot \left( \frac 35 \cdot \frac 23 \right)^x = \frac {40}{225}$

$\Rightarrow \left( \frac 25 \right)^x = \frac 8{75}$
• Feb 5th 2009, 09:01 PM
Logs
Hello Shandy
Quote:

Originally Posted by Shandy
If someone could help me to solve this, I would appreciate it a lot.

(3/5)^x-1 * (2/3)^x = 40/225

$\left(\frac{3}{5}\right)^{x-1}\times\left(\frac{2}{3}\right)^{x} = \frac{40}{225}$

Take logs of both sides, and use the laws:

$\log(a^b) = b\log (a)$ and $\log(a\times b) = \log(a) + \log(b)$

and get:

$(x-1) \log \left(\frac{3}{5}\right) + x\log\left(\frac{2}{3}\right) = \log\left(\frac{40}{225}\right)$

Collect together the like terms:

$x\left( \log \left(\frac{3}{5}\right) + \log\left(\frac{2}{3}\right)\right) = \log\left(\frac{40}{225}\right)+ \log \left(\frac{3}{5}\right)$

$\Rightarrow x\log\left(\frac{3\times 2}{5 \times 3}\right) = \log\left(\frac{40\times 3}{225 \times 5} \right)$

$\Rightarrow x\log\left(\frac{2}{5}\right) = \log\left(\frac{8}{75}\right)$

$\Rightarrow x = 2.4425$