algebra twists

**grgrsanjay** · August 18th 2010, 03:29 AM

i m not ans....

find the value for x where $(2.4)^x = (2.6)^{(x-1)}$ has
1. no solution

2. exactly one solution

3. at least two solution

4. infinite number of solution

**Prove It** · August 18th 2010, 04:08 AM

$2.4^x = 2.6^{x-1}$

$\left(\frac{12}{5}\right)^x = \left(\frac{13}{5}\right)^{x - 1}$

$\frac{12^x}{5^x} = \frac{13^{x - 1}}{5^{x-1}}$

$12^x = 5\cdot13^{x-1}$

$\ln{(12^x)} = \ln{(5\cdot13^{x-1})}$

$x\ln{12} = \ln{5} + (x-1)\ln{13}$

$x\ln{12} = \ln{5} + x\ln{13} - \ln{13}$

$\ln{13} - \ln{5} = x\ln{13} - x\ln{12}$

$\ln{13} - \ln{5} = x(\ln{13} - \ln{12})$

$x = \frac{\ln{13} - \ln{5}}{\ln{13} - \ln{12}}$

$x = \frac{\ln{13} - \ln{5}}{\ln{13} - \ln{3} - 2\ln{2}}$ .

**sa-ri-ga-ma** · August 18th 2010, 05:17 AM

$12^x = 13^{x-1}$ should be

$12^x = 5(13^{x-1})$

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