logarithm help

**student4200** · August 25th 2008, 11:53 AM

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

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

**Chop Suey** · August 25th 2008, 12:11 PM

Given: $(4)(3^x) = (7)(5^x)$

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

Property of multiplication in logarithms: $\log{4} + \log{3^x} = \log{7} + \log{5^x}<br />$

Subtract $\log{4}$ from both sides: $\log{3^x} = \log{7} + \log{5^x} - \log{4} <br />$

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

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

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

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

**Jhevon** · August 25th 2008, 12:13 PM

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?

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

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

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

now what?

**Krizalid** · August 25th 2008, 12:57 PM

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

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