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

User Swagatika
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I will demonstrate two different ways to solve the equation 5^(2x+1) = 25.

Method 1: Taking the logarithm of both sides

Step 1: Start with the equation: 5^(2x+1) = 25

Step 2: Take the logarithm of both sides. We can use any logarithm base, but let's use the natural logarithm (ln) for this example:

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

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

(2x+1) * ln(5) = ln(25)

Step 4: Divide both sides by ln(5) to solve for x:

2x+1 = ln(25) / ln(5)

Step 5: Subtract 1 from both sides:

2x = (ln(25) / ln(5)) - 1

Step 6: Divide both sides by 2 to isolate x:

x = ((ln(25) / ln(5)) - 1) / 2

Now you can evaluate the right side of the equation to find the numerical value of x.

Method 2: Using the properties of exponents

Step 1: Start with the equation: 5^(2x+1) = 25

Step 2: Rewrite 25 as a power of 5:

5^2 = 25

Step 3: Set the exponents equal to each other:

2x+1 = 2

Step 4: Subtract 1 from both sides:

2x = 2 - 1

Step 5: Divide both sides by 2 to solve for x:

x = (2 - 1) / 2

Simplifying further:

x = 1/2

So, the solutions for the equation 5^(2x+1) = 25 are x = ((ln(25) / ln(5)) - 1) / 2 and x = 1/2.

User Gondim
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