Find the value of X in the exponential equation. 8^(2x-6) = 16^(2x-3).x = 2. Find the value of X in the exponential equation 1/16 = 8^(x-3). x = (Answer should be in a fraction form, i.e 1/2)
The Correct Answer and Explanation is :
Let’s solve both exponential equations step by step.
Problem 1:
[
8^{(2x-6)} = 16^{(2x-3)}
]
Step 1: Express the bases as powers of 2
Since:
[
8 = 2^3 \quad \text{and} \quad 16 = 2^4
]
We rewrite the equation as:
[
(2^3)^{(2x-6)} = (2^4)^{(2x-3)}
]
Step 2: Apply the Power Rule ((a^m)^n = a^{m \cdot n})
[
2^{3(2x-6)} = 2^{4(2x-3)}
]
[
2^{(6x-18)} = 2^{(8x-12)}
]
Step 3: Set Exponents Equal to Each Other
Since the bases are the same, we equate the exponents:
[
6x – 18 = 8x – 12
]
Step 4: Solve for (x)
[
-18 + 12 = 8x – 6x
]
[
-6 = 2x
]
[
x = -\frac{6}{2} = -3
]
Answer for Problem 1:
[
x = -3
]
Problem 2:
[
\frac{1}{16} = 8^{(x-3)}
]
Step 1: Express the Bases as Powers of 2
We rewrite ( \frac{1}{16} ) and ( 8 ) using base 2:
[
\frac{1}{16} = 2^{-4} \quad \text{and} \quad 8 = 2^3
]
Thus, we rewrite the equation as:
[
2^{-4} = (2^3)^{(x-3)}
]
Step 2: Apply the Power Rule ((a^m)^n = a^{m \cdot n})
[
2^{-4} = 2^{3(x-3)}
]
Step 3: Set Exponents Equal to Each Other
Since the bases are the same, we equate the exponents:
[
-4 = 3(x – 3)
]
Step 4: Solve for (x)
[
-4 = 3x – 9
]
[
-4 + 9 = 3x
]
[
5 = 3x
]
[
x = \frac{5}{3}
]
Answer for Problem 2:
[
x = \frac{5}{3}
]
Explanation (300 words)
In both problems, we solve for (x) in an exponential equation by expressing the given bases as powers of a common base, typically 2, and then equating the exponents.
For the first equation, we rewrite (8) and (16) as powers of 2, leading to an equation where exponents are equal. Solving this linear equation gives (x = -3).
For the second equation, rewriting (1/16) and (8) as base-2 expressions allows us to set the exponents equal, leading to another linear equation. Solving it results in (x = \frac{5}{3}).
These techniques are useful in algebra and calculus for solving exponential and logarithmic equations efficiently.