# Why is 10^-x = 5^4x equal to -x log 10 = 4x log 5 when ...

• Dec 13th 2012, 05:54 PM
LyndaCS
Why is 10^-x = 5^4x equal to -x log 10 = 4x log 5 when ...
I'm having trouble with the right side of this problem: 10^-x = 5^4x. Why is 10^-x = 5^4x processed to -x log 10 = 4x log 5? The right side means 4x log (base 10) 5, right? I'm confused because the 5 on the right side of the original problem is that side's base. Why wouldn't it be: -x log 10 = 4x log base 5? Seems like in some problems the number in that spot, ie the 5, is used as the log base and sometimes it isn't, and I don't know when or why. Please help, as the final's in 1.5 days!
• Dec 13th 2012, 06:18 PM
MarkFL
Re: Why is 10^-x = 5^4x equal to -x log 10 = 4x log 5 when ...
If you take logs of both sides of an equation, you have to use the same base, otherwise you no longer have an equation, since you have done unequal operations to the two sides of the equation. You could use base 5 logs, but base 10 is preferred as most calculators have base 10 logs as intrinsic functions.
• Dec 14th 2012, 05:21 AM
LyndaCS
Re: Why is 10^-x = 5^4x equal to -x log 10 = 4x log 5 when ...
OK. The reason I'm confused is that my books says that 2^x = 16 converts to logarithm log_2 16 = x. The 2 is used as the base. But in problem 10^-x = 5^4x , the 5 isn't used as the base. Must be the difference is that the second problem has variables in exponents on both sides so it's handled one way, and when there's only an exponent variable on one side it's handled the other. Is that right?
• Dec 14th 2012, 05:37 AM
MarkFL
Re: Why is 10^-x = 5^4x equal to -x log 10 = 4x log 5 when ...
Yes, when the variable is in an exponent on both sides of the equation, it is usually more convenient to take logs of both sides.