1. ## logarithm help

I need help with this problem. I have trouble using log to isolate x.

(4)(3^x) = (7)(5^x)

2. Given: $\displaystyle (4)(3^x) = (7)(5^x)$

Take the log of both sides: $\displaystyle \log{[(4)(3^x)]} = \log{[(7)(5^x)]}$

Property of multiplication in logarithms: $\displaystyle \log{4} + \log{3^x} = \log{7} + \log{5^x}$

Subtract $\displaystyle \log{4}$ from both sides: $\displaystyle \log{3^x} = \log{7} + \log{5^x} - \log{4}$

Subtract $\displaystyle \log{5^x}$ from both sides: $\displaystyle \log{3^x} - \log{5^x} = \log{7} - \log{4}$

Combine the two logs on the LHS: $\displaystyle \log{\left(\frac{3}{5}\right)^x} = \log{7} - \log{4}$

Carry down x (property of logarithm): $\displaystyle x \log{\left(\frac{3}{5}\right)} = \log{7} - \log{4}$

Divide by $\displaystyle \log{\left(\frac{3}{5}\right)}$ on both sides: $\displaystyle x = \frac{\log{7} - \log{4}}{\log{\left(\frac{3}{5}\right)}}$

3. Originally Posted by student4200
I need help with this problem. I have trouble using log to isolate x.

(4)(3^x) = (7)(5^x)
do you know the rules of logs?

$\displaystyle \log 4(3^x) = \log 7(5^x)$ .................log both sides

$\displaystyle \Rightarrow \log 4 + \log 3^x = \log 7 + \log 5^x$ ...............since $\displaystyle \log_a XY = \log_a X + \log_a Y$

$\displaystyle \Rightarrow \log 4 + x \log 3 = \log 7 + x \log 5$ ............since $\displaystyle \log_a X^n = n \log_a X$

now what?

4. $\displaystyle 4\cdot 3^{x}=7\cdot 5^{x}\implies \left( \frac{3}{5} \right)^{x}=\frac{7}{4}\,\therefore\, x=\frac{\ln 7-\ln 4}{\ln 3-\ln 5}.$