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Sagot :
To determine which exponential function has an [tex]\( x \)[/tex]-intercept, we need to find for which functions [tex]\( f(x) = 0 \)[/tex] has a real solution for [tex]\( x \)[/tex].
Let's analyze each function one by one:
Function A: [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 100^{x-5} - 1 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 100^{x-5} = 1 \][/tex]
The equation [tex]\( 100^{x-5} = 1 \)[/tex] holds true if [tex]\( 100^{x-5} = 100^0 \)[/tex], which happens when:
[tex]\[ x-5 = 0 \implies x = 5 \][/tex]
Thus, function A has a real [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex].
Function B: [tex]\( f(x) = 3^{x-4} + 2 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 3^{x-4} + 2 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 3^{x-4} = -2 \][/tex]
The base of the exponent, 3, is a positive number, and any power of a positive number remains positive. Hence, [tex]\( 3^{x-4} = -2 \)[/tex] has no real solution because a positive number can never equal a negative number.
Thus, function B has no real [tex]\( x \)[/tex]-intercept.
Function C: [tex]\( f(x) = 7^{x-1} + 1 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 7^{x-1} + 1 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 7^{x-1} = -1 \][/tex]
Similarly to function B, the base 7 is positive, and powers of positive numbers cannot equal negative numbers.
Thus, function C has no real [tex]\( x \)[/tex]-intercept.
Function D: [tex]\( f(x) = -8^{x+1} - 3 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -8^{x+1} - 3 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ -8^{x+1} = 3 \][/tex]
Here, [tex]\( -8^{x+1} \)[/tex] indicates a negative base raised to a power, which will always yield negative results, and therefore cannot be equal to a positive number.
Thus, function D has no real [tex]\( x \)[/tex]-intercept.
Conclusion:
By examining all functions, we see that the only function which has a real [tex]\( x \)[/tex]-intercept is function A, [tex]\( f(x) = 100^{x-5} - 1 \)[/tex], at [tex]\( x = 5 \)[/tex].
So, the correct answer is A.
Let's analyze each function one by one:
Function A: [tex]\( f(x) = 100^{x-5} - 1 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 100^{x-5} - 1 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 100^{x-5} = 1 \][/tex]
The equation [tex]\( 100^{x-5} = 1 \)[/tex] holds true if [tex]\( 100^{x-5} = 100^0 \)[/tex], which happens when:
[tex]\[ x-5 = 0 \implies x = 5 \][/tex]
Thus, function A has a real [tex]\( x \)[/tex]-intercept at [tex]\( x = 5 \)[/tex].
Function B: [tex]\( f(x) = 3^{x-4} + 2 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 3^{x-4} + 2 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 3^{x-4} = -2 \][/tex]
The base of the exponent, 3, is a positive number, and any power of a positive number remains positive. Hence, [tex]\( 3^{x-4} = -2 \)[/tex] has no real solution because a positive number can never equal a negative number.
Thus, function B has no real [tex]\( x \)[/tex]-intercept.
Function C: [tex]\( f(x) = 7^{x-1} + 1 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 7^{x-1} + 1 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ 7^{x-1} = -1 \][/tex]
Similarly to function B, the base 7 is positive, and powers of positive numbers cannot equal negative numbers.
Thus, function C has no real [tex]\( x \)[/tex]-intercept.
Function D: [tex]\( f(x) = -8^{x+1} - 3 \)[/tex]
1. Set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ -8^{x+1} - 3 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ -8^{x+1} = 3 \][/tex]
Here, [tex]\( -8^{x+1} \)[/tex] indicates a negative base raised to a power, which will always yield negative results, and therefore cannot be equal to a positive number.
Thus, function D has no real [tex]\( x \)[/tex]-intercept.
Conclusion:
By examining all functions, we see that the only function which has a real [tex]\( x \)[/tex]-intercept is function A, [tex]\( f(x) = 100^{x-5} - 1 \)[/tex], at [tex]\( x = 5 \)[/tex].
So, the correct answer is A.
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