SOLUTION: could sombody help me to solve it? 1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: could sombody help me to solve it? 1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))      Log On


   

Question 202664: could sombody help me to solve it?
1/3.log(20,2x-1)=log(20,11)-2.log(20,(2x-1)^(1/3))+log(5,(log7,7))
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Quick note: log%287%2C%287%29%29=1 and log%285%2C%281%29%29=0. So log%285%2C%28log%287%2C%287%29%29%29%29=log%285%2C%281%29%29=0. In other words, log%285%2C%28log%287%2C%287%29%29%29%29=0


So the expression


simplifies to


-----------------------------------



Start with the given expression.


Rewrite the last log using the identity log%28b%2C%28x%5Ey%29%29=y%2Alog%28b%2C%28x%29%29


Multiply


Subtract log%2820%2C%2811%29%29 from both sides.


Subtract %281%2F3%29log%2820%2C%282x-1%29%29 from both sides.


-log%2820%2C%2811%29%29=%28-2%2F3-1%2F3%29log%2820%2C%282x-1%29%29 Factor out the GCF log%2820%2C%282x-1%29%29


-log%2820%2C%2811%29%29=%28-3%2F3%29log%2820%2C%282x-1%29%29 Combine like terms.


-log%2820%2C%2811%29%29=-log%2820%2C%282x-1%29%29 Reduce


log%2820%2C%2811%29%29=log%2820%2C%282x-1%29%29 Multiply both sides by -1


11=2x-1 Since the bases of the logs are equal, this means that the arguments are equal.


11%2B1=2x Add 1 to both sides.


12=2x Combine like terms.


12%2F2=x Divide both sides by 2 to isolate "x".


6=x Reduce


x=6 Rearrange the equation.


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Answer:


So the solution is x=6