Question
Use the properties of exponents to rewrite the function f(x)=351−x+7
in the form f(x)=a(b)x−h+k
.
in the form f(x)=a(b)x−h+k
.
Answers
Answered by
GPT-4o mini
To rewrite the function \( f(x) = 351 - x + 7 \) in the desired form \( f(x) = a(b)^{x - h} + k \), we first need to simplify the expression.
1. Combine the constants in the function:
\[
f(x) = 351 - x + 7 = 358 - x
\]
Next, we want to express \( 358 - x \) in the form \( a(b)^{x-h} + k \).
To rewrite \( 358 - x \), we can factor the negative sign in front of \( x \):
2. Rewrite \( 358 - x \):
\[
f(x) = -x + 358 = -1(x - 358)
\]
However, this doesn't quite fit the exponential form we want.
To express it in an exponential form, we can do the following:
3. Recognize that a linear function like \( -x + 358 \) does not easily convert directly into a purely exponential form, but if we think of a decay-like function, we can write:
\[
f(x) = 358 - 1 \cdot e^{-x + \ln(358)}
\]
Noticing \( e^{\ln(358)} = 358 \).
This can be a crude way of expressing it, or we can consider simply adjusting our algorithm:
Consider some transformed exponential base by choosing \( b = e \) and shift transformations while keeping constant values outside the base.
In general terms \( a = 358 \), \( b = e^{-1} \), then we'd rewrite:
\[
f(x) = -\left(\frac{1}{e^{x - \ln(358)}}\right) + k
\]
4. To give precise values:
\[
f(x) = 358 - x \; \text{can also loosely be expressed in accompanying exponential forms, however typically a linear will remain that way.}
\]
### Conclusion
Thus the direct conversion to \( a(b)^{x-h} + k \) is somewhat restrictive or unconventional, given a proper \( a, b \) identification:
If you accept transforms:
\[
f(x) = 358 - x \text{ with linear inputs }, \text{ however feel the framework does not perfectly fit into an exponential compression}.
\]
The function maintains traditional linearity; functions of that construction do not convert well into pure exponential without specific transformations obscured.
Should you aim to fit strictly:
\[
f(x) = 358 \cdot (0)^{x - 1} + 358 \text{ serves as positioning whereas isolating exponentials returns elegantly trivial or infeasible.}
\]
Yeah, simple outcomes underscore their independent definitions serve more accuracy than forcing arbitrary correspondences.
1. Combine the constants in the function:
\[
f(x) = 351 - x + 7 = 358 - x
\]
Next, we want to express \( 358 - x \) in the form \( a(b)^{x-h} + k \).
To rewrite \( 358 - x \), we can factor the negative sign in front of \( x \):
2. Rewrite \( 358 - x \):
\[
f(x) = -x + 358 = -1(x - 358)
\]
However, this doesn't quite fit the exponential form we want.
To express it in an exponential form, we can do the following:
3. Recognize that a linear function like \( -x + 358 \) does not easily convert directly into a purely exponential form, but if we think of a decay-like function, we can write:
\[
f(x) = 358 - 1 \cdot e^{-x + \ln(358)}
\]
Noticing \( e^{\ln(358)} = 358 \).
This can be a crude way of expressing it, or we can consider simply adjusting our algorithm:
Consider some transformed exponential base by choosing \( b = e \) and shift transformations while keeping constant values outside the base.
In general terms \( a = 358 \), \( b = e^{-1} \), then we'd rewrite:
\[
f(x) = -\left(\frac{1}{e^{x - \ln(358)}}\right) + k
\]
4. To give precise values:
\[
f(x) = 358 - x \; \text{can also loosely be expressed in accompanying exponential forms, however typically a linear will remain that way.}
\]
### Conclusion
Thus the direct conversion to \( a(b)^{x-h} + k \) is somewhat restrictive or unconventional, given a proper \( a, b \) identification:
If you accept transforms:
\[
f(x) = 358 - x \text{ with linear inputs }, \text{ however feel the framework does not perfectly fit into an exponential compression}.
\]
The function maintains traditional linearity; functions of that construction do not convert well into pure exponential without specific transformations obscured.
Should you aim to fit strictly:
\[
f(x) = 358 \cdot (0)^{x - 1} + 358 \text{ serves as positioning whereas isolating exponentials returns elegantly trivial or infeasible.}
\]
Yeah, simple outcomes underscore their independent definitions serve more accuracy than forcing arbitrary correspondences.