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 ->  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

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 Click here to see ALL problems on logarithm 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(28595)   (Show Source): You can put this solution on YOUR website!Quick note: and . So . In other words, So the expression simplifies to ----------------------------------- Start with the given expression. Rewrite the last log using the identity Multiply Subtract from both sides. Subtract from both sides. Factor out the GCF Combine like terms. Reduce Multiply both sides by -1 Since the bases of the logs are equal, this means that the arguments are equal. Add 1 to both sides. Combine like terms. Divide both sides by 2 to isolate "x". Reduce Rearrange the equation. ================================================================== Answer: So the solution is