# Explain how i determine roots in equation 4+2/((5^x)-1)=3/(5^(x-1))?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use first the following exponent rule for the denominator of fraction, to the right side, such that:

`a^(x - y) = (a^x)/(a^y)`

Reasoning by analogy, yields:

`5^(x - 1) = 5^x/5`

Replacing `5^x/5` for `5^(x - 1)` yields:

`4 + 2/(5^x - 1) = 3/(5^x/5) => 4 + 2/(5^x - 1) = 15/5^x`

You need to bring the terms to a common denominator, such that:

`4*5^x*(5^x - 1) + 2*5^x = 15*(5^x - 1)`

`4*5^(2x) - 4*5^x + 2*5^x - 15*5^x + 15 = 0`

`4*5^(2x) - 17*5^x + 15 = 0`

You should come up with the following substitution, such that:

`5^x = y`

Replacing the variable in equation, yields:

`4y^2 - 17y + 15 = 0`

`y_(1,2) = (17+-sqrt(289 - 240))/8`

`y_(1,2) = (17+-sqrt49)/8`

`y_(1,2) = (17+-7)/8 => y_1 = 3 ; y_2 = 5/4`

You need to solve the following equations, such that:

`5^x = y_1 => 5^x = 3 `

Taking common logarithms yields:

`log 5^x = log 3 => x*log 5 = log 3 => x = log 3/log 5`

`5^x = y_2 => 5^x = 5/4`

Taking common logarithms yields:

`x*log 5 = log (5/4) => x = (log (5/4))/(log 5)`

`x = (log 5 - log 4)/(log 5) => x = log 5/log 5 - log 4/log 5 => x = 1 - log 4/log 5` .

Hence, evaluating the solutions to the given equation, yields `x = log 3/log 5` , `x = 1 - log 4/log 5.`