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Demonstrate at least two different ways how to solve the equation 5^(2x +1)=25

User Cdmt
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1 Answer

3 votes

Answer:

x = 1/2

Explanation:

1) Using logarithms

5^(2x +1) = 25

log₅(5^(2x +1)) = log₅(25) . . . . take log base 5

2x +1 = 2 . . . . . . . . . . . . . . . . simplify

2x = 1 . . . . . . . . . . . . . . . . . subtract 1

x = 1/2 . . . . . . . . . . . . divide by 2

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2) Equating exponents (basically, the same thing)

5^(2x +1) = 25

(5^(2x))(5^1) = 25

5^(2x) = 25/5 = 5^1

2x = 1 . . . . . equate exponents

x = 1/2

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3) See attached for a graphical solution to ...

f(x) = 5^(2x+1) -25

for f(x) = 0

The x-intercept is shown as x = 0.5.

Demonstrate at least two different ways how to solve the equation 5^(2x +1)=25-example-1
User Nextar
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5.9k points