SOLUTION: solve 3log_(5)y-log

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Question 1077085: solve 3log_(5)y-log_(y)5=2
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
this winds up being able to solve a quadratic equation.
the math is horrendous, but i used an online quadratic equation calculator to solve to the degree of accuracy required.

the concept is as follows:

start with 3log5(y) - logy(5) = 2

conver to base of 10 log as follows:

log5(y) = log(6)/log(5)

logy(5) = log(5)/log(y)

equation becomes 3log(y)/log(5) - log(5)/log(y) = 2

log(anything) is assumeed to be log10(anything).

your calculator can handle log10(anything) by use of the calculator LOG function.

so your equation has become:

3log(y)/log(5) - log(5)/log(y) = 2

multiply both sides of this equation by log(5)log(y) to get:

3log(y)^2 - log(5)^2 = 2log(5)log(y)

subtract 2log(5)log(y) from both sides of the equation to get:

3log(y)^2 - 2log(5)log(y) - log(5)^2 = 0

let x = log(y).

equation becomes 3x^2 - 2log(5)x - log(5)^2 = 0

this is a quadratic equation that can be solved by using the quadratic formula.

i spared myself the drudgery by using the following online quadratic equation solver:

https://www.mathsisfun.com/quadratic-equation-solver.html

since this solver couldn't handle -2log(5) or -log(5)^2, i had to convert them to fractions.

-2log(5) became -1.397940009

-log(5)^2 became -.48859067

these were input into the quadratic solve and the result was:

x = 0.69897000443178 or x = -0.23299000143178

since x is equal to log(y), then the solution became:

log(y) = 0.69897000443178 or log(y) = -0.23299000143178

log(y) = 0.69897000443178 if and only if 10^0.69897000443178 = y

this led to y = 5.000000001 or y = .5848035477

it remained to place these values into the original equation to see if that equation holds true.

it holds true for y = 5.000000001

it also holds true for y = .5848035477

the solution from the quadratic solver is shown below:

i also solved the equation graphically as shown below:

the graphical solutions are rounded to 3 decimal places but they confirm the alebraic solutuion is correct.

in the graph, i used x instead of y.

same equation with a change in variable name.

the intersection of the 2 graphs is the solution, with the correct value of the original y being the x-coordinate of the coordinate pair.

for example, the coordinate pair of (.585,2) says that the value of x is .585.

you would simply read that as y = .585.

similarly, the coordinate pair of (5,2) says that the value of x is 5.

you would simply read that as y = 5.

SOLUTION: solve 3log_(5)y-log_(y)5=2