Solve for x:  25^x - 6*5^x + 5 = 0 using log or ln.

Expert Answers

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You should come up with the substitution `5^x = y`  such that:

`5^(2x) - 6*5^x + 5 = 0`

`y^2 - 6y + 5 = 0`

You should use quadratic formula such that:

`y_(1,2) = (-(-6)+-sqrt((-6)^2 - 4*1*5))/2`

`y_(1,2) = (6 +- sqrt(36 - 20))/2`

`y_(1,2) = (6 +- sqrt16)/2`

`y_(1,2) = (6 +- 4)/2`

`y_1 = (6+4)/2 =gt y_1 = 5`

`y_2 = (6-4)/2 =gt y_2 = 1`

You should solve for x the equations `5^x = 5`  and `5^x = 1`  such that:

`5^x = 5`

Using logarithms yields:

`log 5^x = log 5 =gt x*log 5 = log 5`

Dividing by log 5 yields:

`x = 1`

`5^x = 1`

You need to take logarithms both sides such that:

`log 5^x = log 1 =gt x*log 5 = 0 =gt x = 0`

Hence, evaluating solutions to equation yields `x = 0`  and `x = 1.`

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