solve for x in the expression below:

Question

5^{4x -1} = 7^{x + 2}

Answer 1

Christopher,

 

As I understand it the problem is to solve 5^4x-1 = 7^x+2 for x.  The + is a bit difficult to see on my screen and looks like it could be a -, so if I am getting that sign wrong, you can change my solution so it still will work.

 

When you see an exponential like 5^4x-1 in an equation, remember that exponentials and logarithms are inverse functions.   

 

If the equation you gave is true, this means that, using any base for the log,

 log(5^4x-1) = log(7^x+2).  

 

However the rule log(a^b)  = b*log(a) means that the equation can be reduced to

(4x-1)*log(5) = (x+2)*log(7).

 

Now you have a simple equation in x, so it becomes 4xlog(5) - log 5 = xlog(7) + 2log(7), and finally, 

x(4*log(5) - log(7)) = log(5) + 2log(7), and so x = (log(5) + 2log(7)) / (4*log(5) - log(7)).

 

This is true for any base log, so you can use whatever base you want to calculate - base 10, base 2, or natural log (base e) - whatever base of log that your calculator can do.

 

So why is it not important what the base of the log is?  It's because of the log base conversion formula

that states that, for any two logarithm bases b and c, log_b(w) = log_c(w)/log_c(b).  Changing the log base multiples the numerator and denominator of the solution expression by the same number and does not change it's value.