Thread: algebra twists

i m not ans....

find the value for x where$\displaystyle (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

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

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

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

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

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

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

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

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

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

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

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

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

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