3Square root of log(5x-1)/(2x-1)=log1/3 x-1, what is the property used to rewrite the equation?

Question

3Square root of log(5x-1)/(2x-1)=log1/3 x-1

 

 

 

STEPS AND ANSWER please and thank you very much for ur help

Answer 1

If I am interpreting your parentheses correctly I believe we have:

 

3√(log(5x-1))/(2x-1) = (log 1)/(3x-1)

 

First... log a = b is understood to be base 10 or lob₁₀(a)=b so 10^b=a

 

Note that log 1 is asking "to what power do I raise 10 to get 1?" ... or 10ⁿ = 1, so n =0

Thus, log 1 = 0  so we have...

 

3√(log(5x-1))/(2x-1) = 0

 

multiply both sides by (2x-1)/3 we have

 

√(log(5x-1)) = 0

 

square both sides and we have

 

log(5x-1) = 0

 

Again, since log a=b means 10^b = a we have...

 

10⁰ = 5x-1

 

1 = 5x-1

 

5x = 2

 

x = 2/5