Please help me solve this equation:
log v27x = 1 - log v27 (x-0.4)
Answer key: x=5.4
***The v means the 27's are lower case.
Thank-you!
log v27x = 1 - log v27 (x-0.4)
Answer key: x=5.4
***The v means the 27's are lower case.
Thank-you!
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Ok first things first, you should know that when subtracting logs your actually dividing whatever is on the inside, and when your adding your multiplying whatever is on the inside (just thought that info. might come in handy as future reference)
Anyways:
log v27x = 1 - log v27 (x-0.4)
logv27x + logv27(x-0.4) = 1 add over the other log, now since we're "adding" we're multiplying
the inside parts
logv27(x^2 - 0.4x) = 1 now put it into exponential form ...
27^1 = (x^2 - 0.4x) its now in exponential form.. so subtract over the 27..
(x^2 - 0.4x - 27) = 0 now use the quadratic formula to solve it ...
[-0.4 +/- √ (0.4)^2 + 4(1)(27)]/ 2(1) you should two answers from this ... (-5) and (5.4)
the reason the answer is ONLY 5.4 is because when you plug in (-5) into x it wont work since when it comes to logarithmic functions the result in the parentheses must be positive
logv27(-5) <------ negative logv27 (-5-0.4) <------- negative in parentheses = BIG nono (a.k.a no solution)
anyways... hope that helped (:
Anyways:
log v27x = 1 - log v27 (x-0.4)
logv27x + logv27(x-0.4) = 1 add over the other log, now since we're "adding" we're multiplying
the inside parts
logv27(x^2 - 0.4x) = 1 now put it into exponential form ...
27^1 = (x^2 - 0.4x) its now in exponential form.. so subtract over the 27..
(x^2 - 0.4x - 27) = 0 now use the quadratic formula to solve it ...
[-0.4 +/- √ (0.4)^2 + 4(1)(27)]/ 2(1) you should two answers from this ... (-5) and (5.4)
the reason the answer is ONLY 5.4 is because when you plug in (-5) into x it wont work since when it comes to logarithmic functions the result in the parentheses must be positive
logv27(-5) <------ negative logv27 (-5-0.4) <------- negative in parentheses = BIG nono (a.k.a no solution)
anyways... hope that helped (: