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Demonstrate two different ways to solve 2^(2x-1) = 25

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User Festus
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Step-by-step explanation:

Method 1: Taking the logarithm of both sides:

Step 1: Start with the equation 2^(2x-1) = 25.

Step 2: Take the logarithm of both sides. We can use any base logarithm; here, we'll use the natural logarithm (ln):

ln(2^(2x-1)) = ln(25).

Step 3: Apply the logarithm property to bring down the exponent:

(2x-1) ln(2) = ln(25).

Step 4: Divide both sides of the equation by ln(2):

2x - 1 = ln(25) / ln(2).

Step 5: Solve for x by isolating the variable:

2x = (ln(25) / ln(2)) + 1.

Step 6: Divide both sides by 2:

x = [(ln(25) / ln(2)) + 1] / 2.

So, one possible solution to the equation 2^(2x-1) = 25 is x = [(ln(25) / ln(2)) + 1] / 2.

Method 2: Solving by manipulating the equation:

Step 1: Start with the equation 2^(2x-1) = 25.

Step 2: Rewrite 25 as 2^2 5:

2^(2x-1) = 2^2 * 5.

Step 3: Apply the power property of equality:

2x - 1 = 2 + log₂(5).

Step 4: Simplify the equation:

2x = 3 + log₂(5).

Step 5: Divide both sides by 2:

x = (3 + log₂(5)) / 2.

So, another possible solution to the equation 2^(2x-1) = 25 is x = (3 + log₂(5)) / 2.

Both methods yield a solution, and you can choose the one that is more convenient or suitable for your particular situation.

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User Ptica
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