Exponents and Logs

Jan 2010
60
0
Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

\(\displaystyle 2(5)^x=7^{x+1}\)
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
Why is the 5 in this equation, I can't multiply the 2 and 5, some of this math is so confusing

\(\displaystyle 2(5)^x=7^{x+1}\)
The 5 is the base of the exponential. You can still take logs and use the relevant log laws.

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

Plato

MHF Helper
Aug 2006
22,461
8,633
\(\displaystyle 2(5)^x=7^{x+1}\)
That equation is equivalent to \(\displaystyle \left( {\frac{5}{7}} \right)^x = \frac{7}{2}\) which has solution \(\displaystyle x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}\)
 
Jan 2010
60
0
That equation is equivalent to \(\displaystyle \left( {\frac{5}{7}} \right)^x = \frac{7}{2}\) which has solution \(\displaystyle x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}\)
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 \(\displaystyle 3^{x+1}\)
 

Plato

MHF Helper
Aug 2006
22,461
8,633
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 \(\displaystyle 3^{x+1}\)
\(\displaystyle 2 \cdot 5^x = 3^{x + 1} \) is equivalent to \(\displaystyle \left( {\frac{5}{3}} \right)^x = \frac{3}{2}\) which has solution \(\displaystyle x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}\)
To see how that works divide both sides by \(\displaystyle 3^x\) and also by \(\displaystyle 2\).
Note that \(\displaystyle \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x \).
 
Jan 2010
60
0
\(\displaystyle 2 \cdot 5^x = 3^{x + 1} \) is equivalent to \(\displaystyle \left( {\frac{5}{3}} \right)^x = \frac{3}{2}\) which has solution \(\displaystyle x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}\)
To see how that works divide both sides by \(\displaystyle 3^x\) and also by \(\displaystyle 2\).
Note that \(\displaystyle \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x \).
Ok, i understand it now, thanks so much for the help, i appreciate it