Exponents and Logs

**nuckers** · May 30th 2010, 02:33 PM

Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

$2(5)^x=7^{x+1}$

**e^(i*pi)** · May 30th 2010, 02:37 PM

Originally Posted by nuckers

Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

$2(5)^x=7^{x+1}$

The 5 is the base of the exponential. You can still take logs and use the relevant log laws.

Hint: $\log(2\cdot 5^x) = \log(2) + \log(5^x)$

**Plato** · May 30th 2010, 03:02 PM

Originally Posted by nuckers

$2(5)^x=7^{x+1}$

That equation is equivalent to $\left( {\frac{5}{7}} \right)^x = \frac{7}{2}$ which has solution $x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}$

**nuckers** · May 30th 2010, 03:19 PM

Originally Posted by Plato

That equation is equivalent to $\left( {\frac{5}{7}} \right)^x = \frac{7}{2}$ which has solution $x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}$

I'm totally confused how you came up with that. I did screw up and put a 7 in the equation, it's supposed to be $3^{x+1}$

**Plato** · May 30th 2010, 03:42 PM

Originally Posted by nuckers

I'm totally confused how you came up with that. I did screw up and put a 7 in the equation, it's supposed to be $3^{x+1}$

$2 \cdot 5^x = 3^{x + 1}$ is equivalent to $\left( {\frac{5}{3}} \right)^x = \frac{3}{2}$ which has solution $x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}$
To see how that works divide both sides by $3^x$ and also by $2$ .
Note that $\frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x$ .

**nuckers** · May 30th 2010, 05:17 PM

Originally Posted by Plato

$2 \cdot 5^x = 3^{x + 1}$ is equivalent to $\left( {\frac{5}{3}} \right)^x = \frac{3}{2}$ which has solution $x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}$
To see how that works divide both sides by $3^x$ and also by $2$ .
Note that $\frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x$ .

Ok, i understand it now, thanks so much for the help, i appreciate it

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