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If
f(x)=2^(5x)+19, what is the value of
f(-1), to the nearest tenth (if necessary)?
Answer:

If $f(x)=2^{5 x}+19$ , what is the value of $f(-1)$ , to the nearest tenth (if necessary)? $\newline$ Answer:

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Powers with decimal bases

Full solution

Q. If

f(x)=2^{5 x}+19

, what is the value of

f(-1)

, to the nearest tenth (if necessary)?

\newline

Answer:

Substitute $x$ with $-1$ : Substitute $x$ with $-1$ in the function $f(x) = 2^{5x} + 19$ . $\newline$ $f(-1) = 2^{5(-1)} + 19$
Calculate exponent part: Calculate the exponent part of the function. $\newline$ $2^{5(-1)} = 2^{-5}$ $\newline$ Since $2^{-5}$ means $1$ over $2$ raised to the power of $5$ , we calculate it as follows: $\newline$ $2^{-5} = 1 / (2^5) = 1 / 32$
Add $19$ : Add $19$ to the result from Step $2$ . $\newline$ $f(-1) = \frac{1}{32} + 19$ $\newline$ To add a fraction to a whole number, we can convert the whole number to a fraction with the same denominator. $\newline$ $19 = 19 \times \left(\frac{32}{32}\right) = \frac{608}{32}$ $\newline$ Now we add the two fractions. $\newline$ $f(-1) = \left(\frac{1}{32}\right) + \left(\frac{608}{32}\right) = \frac{1 + 608}{32} = \frac{609}{32}$
Convert to decimal: Convert the fraction to a decimal to find the nearest tenth. $\newline$ $609 / 32 = 19.03125$ $\newline$ Rounded to the nearest tenth, this is $19.0$ .

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