SOLUTION: I need help solving a problem, i get to a certain point and get stuck. {{{log (5,x)+log (5, 4x-1)=1}}} I get as far as writing the equation in standard formula and the using the qu
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-> SOLUTION: I need help solving a problem, i get to a certain point and get stuck. {{{log (5,x)+log (5, 4x-1)=1}}} I get as far as writing the equation in standard formula and the using the qu
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Question 934389: I need help solving a problem, i get to a certain point and get stuck. I get as far as writing the equation in standard formula and the using the quadratic equation but I think that is where I'm messing up and getting stuck. Found 2 solutions by josgarithmetic, MathTherapy:Answer by josgarithmetic(39618) (Show Source):
, what your actual text really shows inside the rendering tags;
The LOGARITHM is the EXPONENT.
The BASE used is 5.
Change into exponential form.
, which you should have no further trouble to solve.
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The rendering does not always clarify certain things. Your original text was log (5,x)+log (5, 4x-1)=1
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You clearly want to be the expression in a logarithm function. I then included left and right parentheses around the expression, which ensured that the -1 part would appear in the intended position; otherwise, your original text makes the -1 appear to be outside the logarithm function, which would NOT be what you intended. I had planned to ADD 1 to left and right sides of the equation and this would have been in disagreement with your equation.
You can put this solution on YOUR website!
I need help solving a problem, i get to a certain point and get stuck. I get as far as writing the equation in standard formula and the using the quadratic equation but I think that is where I'm messing up and getting stuck.
Was it that difficult to show what quadratic equation you arrived at?
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