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If
f(x)=2^(2x)+10, what is the value of
f(-3), to the nearest ten-thousandth (if necessary)?
Answer:

If $f(x)=2^{2 x}+10$ , what is the value of $f(-3)$ , to the nearest ten-thousandth (if necessary)? $\newline$ Answer:

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Solve exponential equations using logarithms

Full solution

Q. If

f(x)=2^{2 x}+10

, what is the value of

f(-3)

, to the nearest ten-thousandth (if necessary)?

\newline

Answer:

Substitute $x$ with $-3$ : To find the value of $f(-3)$ , we need to substitute $x$ with $-3$ in the function $f(x) = 2^{(2x)} + 10$ . $\newline$ $f(-3) = 2^{(2*(-3))} + 10$
Calculate exponent part: Now we calculate the exponent part: $2\cdot(-3) = -6$ . So we have: $\newline$ $f(-3) = 2^{-6} + 10$
Calculate $2^{-6}$ : Next, we calculate $2^{-6}$ . Since $2^{-6}$ is the same as $1/(2^6)$ , we find: $\newline$ $2^{-6} = 1/(2^6) = 1/64$
Add $10$ : Now we add $10$ to the result of $2^{-6}$ : $\newline$ $f(-3) = \frac{1}{64} + 10$
Express $10$ as fraction: To add $\frac{1}{64}$ to $10$ , we need to express $10$ as a fraction with the same denominator as $\frac{1}{64}$ : $10 = 10 \times \left(\frac{64}{64}\right) = \frac{640}{64}$
Add two fractions: Now we add the two fractions: $f(-3) = \frac{1}{64} + \frac{640}{64} = \frac{(1 + 640)}{64}$
Perform addition in numerator: We perform the addition in the numerator: $f(-3) = (1 + 640)/64 = 641/64$
Divide $641$ by $64$ : Finally, we divide $641$ by $64$ to get the decimal value: $\newline$ $f(-3) = \frac{641}{64} \approx 10.015625$

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