SOLUTION: 5^x − 45(5^−x) + 4 = 0
Solve the equation. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)
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-> SOLUTION: 5^x − 45(5^−x) + 4 = 0
Solve the equation. (Enter your answers as a comma-separated list. Round your answers to four decimal places.)
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Question 1111408: 5^x − 45(5^−x) + 4 = 0
Solve the equation. (Enter your answers as a comma-separated list. Round your answers to four decimal places.) Answer by ikleyn(52798) (Show Source):
Introduce new variable u = .
Then your equation takes the form
u - 45/u + 4 = 0,
u^2 + 4u - 45 = 0,
(u-5)*(u+9) = 0
The last equation has two roots u= 5 and u= -9,
but since u = , only positive "u" works.
Thus u = 5, = 5 and x = 1.
Solved.
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Introducing new variable is the STANDARD method of solving exponential equations.
