Exponential Functions: Additional Practice (Set 2)

Subtract the exponents when dividing like bases.

$\frac{5^8}{5^5} = 5^{8-5} = 5^3 = 125$

The point $(0, 5)$ gives $a = 5$. Then use $(2, 45)$.

$5 b^2 = 45 \;\Rightarrow\; b^2 = 9 \;\Rightarrow\; b = 3$

Write $\tfrac{1}{25}$ as a power of $5$.

$\frac{1}{25} = \frac{1}{5^2} = 5^{-2} \;\Rightarrow\; x = -2$

Multiply by $\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$.

$16 \cdot \frac{9}{4} = 36$

A one-third power is a cube root and a one-half power is a square root.

$27^{1/3} + 16^{1/2} = 3 + 4 = 7$

Adding $3$ to the input multiplies the output by $2^3$.

$f(x+3) = 2^{x+3} = 2^3 \cdot 2^x = 8\,f(x)$

Subtract the exponents.

$\frac{b^{x+3}}{b^x} = b^{(x+3) - x} = b^3$

Write both sides as powers of $3$. Since $\tfrac{1}{3} = 3^{-1}$ and $27 = 3^3$:

$3^{-x} = 3^3 \;\Rightarrow\; -x = 3 \;\Rightarrow\; x = -3$

Split off the added exponent.

$2^{a+3} = 2^a \cdot 2^3 = 5 \cdot 8 = 40$

Multiply exponents for the power of a power, then add for the extra factor.

$\left(2^x\right)^3 \cdot 2 = 2^{3x} \cdot 2^1 = 2^{3x+1}$

At $x = 0$ the output is the base to the zero power.

$f(0) = b^0 = 1$

So the curve passes through $(0, 1)$ for every allowed base.

A negative exponent flips the fraction, then square.

$\left(\tfrac{2}{5}\right)^{-2} = \left(\tfrac{5}{2}\right)^{2} = \frac{25}{4}$