Exponential Functions and Graphs: Additional Practice (Set 2)

Test the small cases, since that is where an exponential can still lose.

$n=1:\; 2 > 1, \qquad n=2:\; 4 = 4, \qquad n=3:\; 8 < 9, \qquad n=4:\; 16 = 16$

Only $n = 3$ satisfies the strict inequality. From $n = 4$ onward the exponential never trails again, because $2^{n} \geq n^{2}$ holds for every integer $n \geq 4$ by induction: doubling the left side beats multiplying the right side by $\left(1 + \tfrac1n\right)^{2} \leq 1.5625$.

Both bases are powers of $2$, so rewrite each side over the base $2$.

$8^{\,x-1} = \left(2^{3}\right)^{x-1} = 2^{\,3x-3}, \qquad 16^{\,x+1} = \left(2^{4}\right)^{x+1} = 2^{\,4x+4}$

Equating exponents, which is legal because an exponential is one-to-one, gives

$3x - 3 = 4x + 4 \quad \Longrightarrow \quad x = -7$

Check: $8^{-8} = 2^{-24}$ and $16^{-6} = 2^{-24}$.

Set $x = 0$ and evaluate the negative exponent as a reciprocal.

$h(0) = 2^{-3} + 1 = \frac{1}{8} + 1 = \frac{9}{8}$

The intercept is $\left(0, \tfrac98\right)$. The value $y = 1$ is the horizontal asymptote here, not the intercept.

Each block of $3$ hours contributes one doubling, so after $t$ hours the count is $200 \cdot 2^{t/3}$.

$200 \cdot 2^{12/3} = 200 \cdot 2^{4} = 200 \cdot 16 = 3200$

Twelve hours is four doublings: $200 \to 400 \to 800 \to 1600 \to 3200$. The distractor $2400$ comes from adding $200$ per doubling instead of multiplying.

Write both factors as powers of $2$ and add the exponents.

$4 \cdot \left(\tfrac{1}{2}\right)^{x} = 2^{2} \cdot 2^{-x} = 2^{\,2-x}$

Check at $x = 0$: the original gives $4 \cdot 1 = 4$, and $2^{2-0} = 4$. At $x = 3$ the original gives $4 \cdot \tfrac18 = \tfrac12$, and $2^{2-3} = 2^{-1} = \tfrac12$.

Write the fraction as a power of $3$.

$\left(\tfrac{1}{9}\right)^{x} = \left(3^{-2}\right)^{x} = 3^{-2x}$

The equation becomes $3^{-2x} = 3^{\,x+6}$, so the exponents must match.

$-2x = x + 6 \quad \Longrightarrow \quad -3x = 6 \quad \Longrightarrow \quad x = -2$

Check: $\left(\tfrac19\right)^{-2} = 81$ and $3^{4} = 81$.

Reflecting across the $x$-axis negates every output, so the new rule is $y = -a \cdot b^{x}$.

$a \cdot b^{x} > 0 \quad \Longrightarrow \quad -a \cdot b^{x} < 0$

The original range was all positive numbers, so the reflected range is all negative numbers. Zero is still excluded, because an exponential is never zero, which rules out the option that includes $y = 0$.

Eighteen days is three half-lives, since $18 \div 6 = 3$, and each half-life multiplies the amount by $\tfrac12$.

$\left(\frac{1}{2}\right)^{3} = \frac{1}{8}$

Decay multiplies rather than subtracts, so three half-lives leave one eighth, not one third. The remaining fraction never reaches zero, which is the asymptote at work.

Put the left side over a single base by rewriting $4$ as $2^{2}$, then add exponents.

$2^{3x} \cdot 4^{x} = 2^{3x} \cdot 2^{2x} = 2^{5x}$

The right side is $32 = 2^{5}$, so $2^{5x} = 2^{5}$, and one-to-oneness gives $5x = 5$, that is $x = 1$.

Check: $2^{3} \cdot 4^{1} = 8 \cdot 4 = 32$.

You never need to know $a$ itself. Split the exponent with the law of exponents.

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

Adding $3$ to the input multiplies the output by $2^{3} = 8$, which is the constant-ratio property at work. Adding $3$ to the exponent does not add $3$ to the value.

A fractional exponent is a root, so name the value exactly before estimating it.

$3^{3/2} = \left(\sqrt{3}\right)^{3} = 3\sqrt{3}$

Since $\sqrt{3} \approx 1.732$, the value is about $3 \cdot 1.732 = 5.196$, which sits between $5$ and $6$. As a check without decimals, $\left(3^{3/2}\right)^{2} = 3^{3} = 27$, and $27$ lies between $5^{2} = 25$ and $6^{2} = 36$.

The inputs $2$, $4$, $6$ are equally spaced, so the outputs have a constant ratio $r = b^{2}$ from one to the next. Over the two steps from $2$ to $6$,

$r^{2} = \frac{320}{20} = 16 \quad \Longrightarrow \quad r = 4.$

So $f(4) = f(2) \cdot r = 20 \cdot 4 = 80$. Equivalently, the middle output is the geometric mean of its neighbours, $\sqrt{20 \cdot 320} = \sqrt{6400} = 80$. The distractor $170$ is their average, which would be correct for a linear function.