Answer:
B. 64 = 2⁽³ˣ⁻⁶⁾.
Step-by-step explanation:
5⁽²ˣ⁻⁵⁾ = 125.
5⁽²ˣ⁻⁵⁾ = 5³
Since exponentials are equal then their powers also will be equal.
2x - 5 = 3
2x = 3 + 5 = 8
x = 8/2
x = 4.
Considering each options:
Option A: 3⁽⁵ˣ⁻¹⁾ = 81.
3⁽⁵ˣ⁻¹⁾ = 3⁴, x = 1 therefore not same solution.
Option B: 64 = 2⁽³ˣ⁻⁶⁾.
2⁽³ˣ⁻⁶⁾ = 2⁶,
3x - 6 = 6
3x = 6 + 6 = 12
x = 12 / 3
x = 4 same solution.
Option C: 4⁽⁷ˣ⁺¹¹⁾ = 256.
4⁽⁷ˣ⁺¹¹⁾ = 4⁴
7x + 11 = 4, x = - 1 therefore not same solution.
Option D: 343 = 7⁽²ˣ⁺⁹⁾.
7⁽²ˣ⁺⁹⁾ = 7³
2x + 9 = 3
2x = - 6
x = -3 therefore not same solution.
Answer:
[tex]\textsf{B. }64=2^{(3x-6)}[/tex]
Step-by-step explanation:
You want the equation with the same solution as ...
[tex]5^{(2x-5)}=125[/tex]
We find it convenient to take the logarithm of the equation and write it in the form f(x) = 0. For the given equation, that is ...
[tex](2x-5)\log(5)=\log(125)\\\\2x-5-\dfrac{\log(125)}{\log(5)}=0[/tex]
This last form is a linear equation that has solution x = 4:
2x -5 -3 = 0
2x = 8
x = 4
The graph of the linear equation is a straight line that crosses the x-axis at x=4.
The given equation and the answer choices are shown graphed in this fashion in the attachment. The only equation with the same x-intercept is ...
[tex]\boxed{\textsf{B. }64=2^{(3x-6)}}[/tex]
__
Additional comment
The values of x that satisfy the equations are ...
Given: x = 4
A. x = 1
B. x = 4 . . . . matches given
C. x = -1
D. x = -3
Of course, we could simply graph 5⁽²ˣ⁻⁵⁾ -125 = 0, which will also have an x-intercept of x=4. This sort of graph has a nearly vertical crossing of the x-axis. The graph of 2⁽³ˣ⁻⁶⁾ -64 = 0 also has a nearly vertical crossing of the x-axis at x=4, so is difficult to differentiate from the graph of the given equation.
We wanted to show the solutions were identical (x=4) without having graphs that could not be differentiated one from the other. If you're going to solve these equations "by hand", the method you would use is that of logarithms as shown. (Actually, you would write the constant as a power of the same base, and equate exponents—the same thing as using logs.)
[tex]5^{2x-5}=5^3\quad\Longrightarrow\quad 2x-5=3[/tex]
Alternatively, rather than solve each of the equations, you could use x=4 in each to see if that makes the equation true. For example, ...
[tex]3^{(5\cdot4-1)}\ne81[/tex]
