What property is used to rewrite this problem? It can't be solved but I want to know how you can...

Question

. 3   ___

log\/5x-1.   =log1

  2x-1.               3  x-1

 

Remember this is a made up question

Answer 1

I'll assume this is your equation:

 

3(√ (log((5x-1)/(2x-1))) ) = log(1/(3x-1))

 

There are a couple things you can do here.

 

 

 

First, divide both sides by 3:

 

√ (log((5x-1)/(2x-1))) = (1/3)log(1/(3x-1))

 

Now you can use the property b(log(a)) = log(a^b) to simplify the right-hand side:

 

√ (log((5x-1)/(2x-1))) = log((1/(3x-1))^1/3)

 

You can simplify further using the property (a/b)^c = a^c/b^c, distributing the 1/3. Remember that 1 to any power is just 1:

 

√ (log((5x-1)/(2x-1))) = log(1/(3x-1)^1/3)

 

 

 

Next, square both sides:

 

log((5x-1)/(2x-1)) = log(1/(3x-1)^1/3)log(1/(3x-1)^1/3)

 

Re-write the equation as exponents of 10 (that is, take 10 to the power of both sides):

 

10^{log((5x-1)/(2x-1))} = 10^{log(1/(3x-1)^(1/3))log(1/(3x-1)^(1/3))}

 

Use the property that a^(log_a(x)) = x to re-write the left-hand side. Use the property that a^bc = (a^b)^c to re-write the right-hand side:

 

(5x-1)/(2x-1) = (10^{log(1/(3x-1)^(1/3))})^{log(1/(3x-1)^(1/3))}

 

Use the property that a^(log_a(x)) = x again to re-write the right-hand side:

 

(5x-1)/(2x-1) = (1/(3x-1)^1/3)^{log(1/(3x-1)^(1/3))}

 

 

 

You can keep re-arranging the equation (ie., distribute the exponent and combine it with 1/3 on the right-hand side), but that should give you an idea of what properties might be useful to you when working with logarithms.

 

You can also plug one of the intermediate steps into Wolfram Alpha to get an actual solution for your equation. It turns out x ≈ 27.0583:

 

http://www.wolframalpha.com/input/?i=log%28%285x-1%29%2F%282x-1%29%29+%3D+log%281%2F%283x-1%29%5E%281%2F3%29%29log%281%2F%283x-1%29%5E%281%2F3%29%29&a=%5E_Principal&a=*FunClash.log-_*Log10-