Exponential Functions and Graphs: Practice

Both $9$ and $27$ are powers of $3$, so rewrite each side over the base $3$.

$9^{\,x+1} = \left(3^{2}\right)^{x+1} = 3^{\,2x+2}, \qquad 27^{\,x-1} = \left(3^{3}\right)^{x-1} = 3^{\,3x-3}$

An exponential is one-to-one, so the exponents must be equal.

$2x + 2 = 3x - 3 \quad \Longrightarrow \quad x = 5$

Check: $9^{6} = 531441$ and $27^{4} = 531441$.

Write every base as a power of $2$, including the fraction.

$\left(\tfrac{1}{4}\right)^{2x} = \left(2^{-2}\right)^{2x} = 2^{-4x}, \qquad 16^{\,x+3} = \left(2^{4}\right)^{x+3} = 2^{\,4x+12}$

Equating the exponents gives

$-4x = 4x + 12 \quad \Longrightarrow \quad 8x = -12 \quad \Longrightarrow \quad x = -\tfrac{3}{2}$

Check: $\left(\tfrac14\right)^{-3} = 64$ and $16^{3/2} = \left(\sqrt{16}\right)^{3} = 64$.

The bases already match, and an exponential is one-to-one, so the exponents must be equal.

$x^{2} = 3x + 10 \quad \Longrightarrow \quad x^{2} - 3x - 10 = 0$

Factor the quadratic.

$(x - 5)(x + 2) = 0 \quad \Longrightarrow \quad x = 5 \; \text{ or } \; x = -2$

Both work, since both make the two exponents agree: $5^{25} = 5^{25}$ and $5^{4} = 5^{4}$.

Apply the law with $y = 0$, which leaves the input unchanged.

$f(x) = f(x + 0) = f(x) \cdot f(0)$

Since $f(x) \neq 0$, divide both sides by it.

$1 = f(0)$

So $b^{0} = 1$ is forced by the law of exponents, not chosen as a convention. (If $f$ were allowed to be zero somewhere it would be zero everywhere, which is why that case is excluded.)

Split the exponent with the law of exponents.

$3^{\,x+2} = 3^{x} \cdot 3^{2} = 9 \cdot 3^{x}$

So sliding the curve two units left is indistinguishable from stretching it vertically by a factor of $9$. Only exponential functions have this double identity, and it is a direct consequence of $b^{x+y} = b^{x}b^{y}$.

Substitute the point and clear the negative exponent by taking a reciprocal.

$b^{-2} = 25 \quad \Longrightarrow \quad \frac{1}{b^{2}} = 25 \quad \Longrightarrow \quad b^{2} = \frac{1}{25}$

A base must be positive, so $b = \tfrac15$. That fits the picture: the curve is decaying, and a decaying exponential does take large values at negative inputs, since $\left(\tfrac15\right)^{-2} = 5^{2} = 25$.

The amount remaining after $t$ years is $\left(\tfrac12\right)^{t/4}$, since each block of $4$ years contributes one halving.

$\left(\tfrac{1}{2}\right)^{12/4} = \left(\tfrac{1}{2}\right)^{3} = \frac{1}{8}$

Twelve years is three half-lives, so the amount is halved three times over: $1 \to \tfrac12 \to \tfrac14 \to \tfrac18$.

Recognize the right side as a power of the reciprocal of the base.

$\frac{27}{8} = \left(\frac{3}{2}\right)^{3} = \left[\left(\frac{2}{3}\right)^{-1}\right]^{3} = \left(\frac{2}{3}\right)^{-3}$

Now both sides share the base $\tfrac23$, so the exponents must agree: $x = -3$. A decaying exponential exceeds $1$ only at negative inputs, which is a useful check on the sign.

Rewrite every option over the common base $2$, then compare exponents.

$4^{9} = 2^{18}, \qquad 2^{16} = 2^{16}, \qquad 8^{5} = 2^{15}, \qquad 16^{4} = 2^{16}$

Since $2^{x}$ is increasing, the largest exponent wins, so $4^{9} = 2^{18} = 262144$ is the greatest. (Notice $2^{16}$ and $16^{4}$ are equal to each other, but both are smaller than $4^{9}$.)

Every value of $1^{x}$ is the same number.

$1^{x} = 1 \qquad \text{for every real } x$

That graph is a horizontal line, so it never grows, never decays, and takes the same output for infinitely many inputs. Being far from one-to-one, it has no inverse to undo it, and none of the machinery of this chapter applies to it. It is excluded by convention, unlike a negative base, which is excluded by necessity.

Write $f(x) = a \cdot b^{x}$. The value at $0$ gives $a = 5$, and then

$5b^{2} = 45 \quad \Longrightarrow \quad b^{2} = 9 \quad \Longrightarrow \quad b = 3.$

So $f(1) = 5 \cdot 3 = 15$. There is a shortcut worth remembering: equally spaced inputs give outputs with a constant ratio, so the middle output is the geometric mean of its neighbours, $\sqrt{5 \cdot 45} = \sqrt{225} = 15$. (Averaging them to get $25$ would be right for a linear function, not an exponential one.)

The two curves meet where $x^{2} = 2^{x}$, and for $x \geq 0$ that happens exactly twice.

$x = 2:\; 4 = 4, \qquad x = 4:\; 16 = 16$

Test a point in each region. At $x = 1$, $2^{1} = 2 > 1 = 1^{2}$. At $x = 3$, $3^{2} = 9 > 8 = 2^{3}$, so the parabola leads. At $x = 5$, $2^{5} = 32 > 25 = 5^{2}$.

So the parabola is ahead only strictly between the crossings, on $2 < x < 4$, and past $x = 4$ the exponential leads forever.