Join the IDNLearn.com community and get your questions answered by experts. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.

Solve for [tex]\( x \)[/tex]:

[tex]\[ (5)^{x^2 + 2x + 7} = (125)^{2x + 1} \][/tex]


Sagot :

Certainly! Let's solve the equation step-by-step:

Given the equation:
[tex]\[ 5^{x^2 + 2x + 7} = 125^{2x + 1} \][/tex]

### Step 1: Simplify the Bases

First, we know that [tex]\(125\)[/tex] can be expressed as a power of [tex]\(5\)[/tex]:
[tex]\[ 125 = 5^3 \][/tex]

Thus, we can rewrite the given equation as:
[tex]\[ 5^{x^2 + 2x + 7} = (5^3)^{2x + 1} \][/tex]

### Step 2: Simplify the Exponents

Using the property of exponents [tex]\((a^m)^n = a^{mn}\)[/tex], we can rewrite the right side:
[tex]\[ (5^3)^{2x + 1} = 5^{3(2x + 1)} = 5^{6x + 3} \][/tex]

Now the equation becomes:
[tex]\[ 5^{x^2 + 2x + 7} = 5^{6x + 3} \][/tex]

### Step 3: Set the Exponents Equal

Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ x^2 + 2x + 7 = 6x + 3 \][/tex]

### Step 4: Rearrange and Simplify the Equation

Rearrange the equation to bring all the terms to one side:
[tex]\[ x^2 + 2x + 7 - 6x - 3 = 0 \][/tex]

Combine like terms:
[tex]\[ x^2 - 4x + 4 = 0 \][/tex]

### Step 5: Solve the Quadratic Equation

Now we have a quadratic equation:
[tex]\[ x^2 - 4x + 4 = 0 \][/tex]

We can solve this quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula. In this case, the equation factors nicely:
[tex]\[ (x - 2)^2 = 0 \][/tex]

So, we find:
[tex]\[ x - 2 = 0 \][/tex]

Thus:
[tex]\[ x = 2 \][/tex]

### Conclusion

By solving the equation, we find:
[tex]\[ x = 2 \][/tex]

Therefore, the solution to the equation [tex]\(5^{x^2 + 2x + 7} = 125^{2x + 1}\)[/tex] is:
[tex]\[ x = 2 \][/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions you’re looking for. Thanks for visiting, and see you next time for more reliable information.