Year 11 Methods — Solving Exponential Equations and Inequalities (13D)

Today's lesson

Now that you can graph exponentials, today is about solving them algebraically. By the end of the lesson you should be able to:

Learning intentions

Solve exponential equations by rewriting both sides with the same base, then equating indices
Use prime decomposition (e.g. $4=2^2$, $27=3^3$, $\sqrt{5}=5^{1/2}$) to force a common base
Solve equations that reduce to a quadratic in disguise (substitution $u=a^x$)
Solve simple exponential inequalities $a^{f(x)} \le a^{g(x)}$ using the rules for $a>1$ vs $0<a<1$
Use the CAS to solve equations whose answer is not a "nice" rational (e.g. $5^x=10$)

Part 1 — Same-base method

Key idea. If $a^x = a^y$ (with $a\in\mathbb{R}^+\setminus\{1\}$), then $x=y$. So our entire strategy is:

Rewrite both sides as powers of the same base.
Equate the exponents.
Solve the resulting linear (or polynomial) equation.

📺 Walkthrough: solve $2^x = 4^{x+1}$ by rewriting $4 = 2^2$, then equating indices.

EXAMPLE 1 — basic same base

Solve (a) $2^x=32$ (b) $5^x=625$.

(a) $32=2^5$, so $2^x=2^5$ ⇒ $\boxed{x=5}$.
(b) $625=5^4$, so $5^x=5^4$ ⇒ $\boxed{x=4}$.

EXAMPLE 2 — rewrite the other base

Solve (a) $2^x=4^{x+1}$ (b) $3^x=9^{x-2}$.

(a) $4=2^2$, so $2^x=(2^2)^{x+1}=2^{2x+2}$ ⇒ $x=2x+2$ ⇒ $\boxed{x=-2}$.
(b) $9=3^2$, so $3^x=3^{2(x-2)}=3^{2x-4}$ ⇒ $x=2x-4$ ⇒ $\boxed{x=4}$.

Now you try: $4^{x+1}=8$. Answer: $2^{2x+2}=2^3\Rightarrow 2x+2=3\Rightarrow x=\tfrac{1}{2}$.

EXAMPLE 3 — products of powers

Solve (a) $7\times 7^{x-2}=49^{4-x}$ (b) $25\times 5^x=125^{2-x}$.

(a) LHS $=7^{1+(x-2)}=7^{x-1}$. RHS $=(7^2)^{4-x}=7^{8-2x}$. So $x-1=8-2x$ ⇒ $3x=9$ ⇒ $\boxed{x=3}$.
(b) LHS $=5^2\cdot 5^x=5^{x+2}$. RHS $=(5^3)^{2-x}=5^{6-3x}$. So $x+2=6-3x$ ⇒ $4x=4$ ⇒ $\boxed{x=1}$.

EXAMPLE 4 — reciprocals and surd bases

Solve (a) $9^{3-x}=\dfrac{1}{27^{3x}}$ (b) $25^{2x-1}=\dfrac{1}{\sqrt{5}}$.

(a) LHS $=3^{2(3-x)}=3^{6-2x}$. RHS $=27^{-3x}=3^{-9x}$. So $6-2x=-9x$ ⇒ $7x=-6$ ⇒ $\boxed{x=-\tfrac{6}{7}}$.
(b) LHS $=5^{2(2x-1)}=5^{4x-2}$. RHS $=5^{-1/2}$. So $4x-2=-\tfrac{1}{2}$ ⇒ $4x=\tfrac{3}{2}$ ⇒ $\boxed{x=\tfrac{3}{8}}$.

Practice 1.1 — solve for $x$ using the same-base method.

$3^x=81$
$2^x=\tfrac{1}{8}$
$4^{x+1}=8$
$9^x=27^{x-1}$
$16^x=\tfrac{1}{2}$
$25^{x-1}=125$

a) $3^x=3^4\Rightarrow x=4$
b) $2^x=2^{-3}\Rightarrow x=-3$
c) $2^{2x+2}=2^3\Rightarrow x=\tfrac{1}{2}$
d) $3^{2x}=3^{3x-3}\Rightarrow x=3$
e) $2^{4x}=2^{-1}\Rightarrow x=-\tfrac{1}{4}$
f) $5^{2x-2}=5^3\Rightarrow x=\tfrac{5}{2}$

Part 2 — Quadratics in disguise

Sometimes an equation looks exponential but is really a quadratic in $a^x$. The trick: substitute $u=a^x$, solve the quadratic, then convert back.

📺 Walkthrough: solve $4^x+2^x-20=0$ by letting $u=2^x$ — get $u^2+u-20=0$, factorise, reject negative root.

EXAMPLE 5 — $4^x+2^x-20=0$

Solve $4^x+2^x-20=0$.

$4^x=(2^2)^x=(2^x)^2$, so the equation becomes $(2^x)^2+2^x-20=0$.
Let $u=2^x$ (where $u>0$). The equation is now $u^2+u-20=0$.
Factorise: $(u-4)(u+5)=0$ ⇒ $u=4$ or $u=-5$.
$u=2^x>0$ for all $x$, so reject $u=-5$.
$2^x=4=2^2$ ⇒ $\boxed{x=2}$.

EXAMPLE 6 — $25^x - 23(5^x) - 50 = 0$

Solve $25^x-23(5^x)-50=0$.

$25^x=(5^2)^x=(5^x)^2$. Let $u=5^x$.
$u^2-23u-50=0$ ⇒ $(u-25)(u+2)=0$ ⇒ $u=25$ or $u=-2$.
Reject $u=-2$ (negative). So $5^x=25=5^2$ ⇒ $\boxed{x=2}$.

Now you try: $9^x-4(3^x)+3=0$. Answer: $u=3^x$, $u^2-4u+3=0$, $(u-1)(u-3)=0$, so $3^x=1$ or $3^x=3$ ⇒ $x=0$ or $x=1$.

Part 3 — Using the CAS for awkward bases

When you can't force a same base, you'll need logarithms (Ex 13E/F next lesson) or a CAS.

EXAMPLE 7 — CAS

Solve $5^x=10$, correct to two decimal places.

On the Nspire: solve(5^x = 10, x) gives the exact answer $x=\dfrac{\ln 10}{\ln 5}$.
Press ctrl+enter to convert to a decimal: $x\approx 1.4307$.
To 2 d.p.: $\boxed{x\approx 1.43}$.

Part 4 — Exponential inequalities

Two rules to know — direction depends on the base:

If $a>1$: $a^x > a^y \;\Leftrightarrow\; x>y$ (inequality direction preserved)
If $0<a<1$: $a^x > a^y \;\Leftrightarrow\; x<y$ (inequality direction reversed)

When in doubt: graph both sides on CAS, find the intersection, and read off the inequality.

EXAMPLE 8 — $16^x > 2$

Solve $16^x > 2$.

$16^x=2^{4x}$, so $2^{4x}>2^1$.
Base $2>1$, so direction preserved: $4x>1$ ⇒ $\boxed{x>\tfrac{1}{4}}$.

EXAMPLE 9 — $3^{2x-1}\le 9$

Solve $3^{2x-1}\le 9$.

$9=3^2$, so $3^{2x-1}\le 3^2$.
Base $3>1$: direction preserved. $2x-1\le 2$ ⇒ $2x\le 3$ ⇒ $\boxed{x\le \tfrac{3}{2}}$.

EXAMPLE 10 — base less than 1

Solve $\left(\tfrac{1}{3}\right)^x\ge 9$.

$\left(\tfrac{1}{3}\right)^x=3^{-x}$ and $9=3^2$, so $3^{-x}\ge 3^2$.
Base $3>1$ (after rewriting), so $-x\ge 2$ ⇒ $\boxed{x\le -2}$.
Check: at $x=-2$, $\left(\tfrac{1}{3}\right)^{-2}=9$ ✓; at $x=-3$, $\left(\tfrac{1}{3}\right)^{-3}=27\ge 9$ ✓.

Now you try: $2^{-3x+1}<\tfrac{1}{16}$. Answer: $2^{-3x+1}<2^{-4}\Rightarrow -3x+1<-4 \Rightarrow x>\tfrac{5}{3}$.

Practice 4.1 — solve each inequality.

$2^x > 8$
$5^{x+1}\le 25$
$9^x\ge \tfrac{1}{27}$
$\left(\tfrac{1}{2}\right)^x < 16$
$\left(\tfrac{1}{4}\right)^x > 32$

a) $x>3$
b) $x+1\le 2\Rightarrow x\le 1$
c) $2x\ge -3\Rightarrow x\ge -\tfrac{3}{2}$
d) $2^{-x}<2^4\Rightarrow -x<4 \Rightarrow x>-4$
e) $2^{-2x}>2^5\Rightarrow -2x>5 \Rightarrow x<-\tfrac{5}{2}$

Part 5 — Quick quiz (5 min)

Working program — Cambridge Ex 13D

After the quiz, open Cambridge Chapter 13 and work through Exercise 13D:

Question	Foundation	Standard (set work)	Advanced
Q1 — basic same base	1 (a,c,e)	1*	1*
Q2 — rewrite one side	2 (a,c)	2*	2*
Q3 — products of powers	3 (a)	3*	3*
Q4 — reciprocals / surds	4 (a)	4*	4*
Q5 — quadratic in $a^x$	—	5*	5*
Q6 — inequalities	6 (a,c)	6*	6*

Next lesson: Ex 13E — introducing logarithms.

Exit ticket — write in your book

Before you pack up, in your exercise book write one sentence each:

What rule lets us "equate indices"? When is it allowed?
In $4^x+2^x-20=0$, why is the substitution $u=2^x$ useful?
If $0<a<1$ and $a^x > a^y$, which way does the inequality between $x$ and $y$ go?