Topic 2.13 Notes – Exponential and Logarithmic Equations and Inequalities

Solving Exponential and Logarithmic Equations

The central inverse relationship is

$$\log_b(x)=c \iff b^c=x$$

with $b>0$, $b\ne1$, and $x>0$.

That gives you the two matching rules you use all the time:

• If $b^u=b^v$, then $u=v$, but only after the bases match.
• If $\log_b(M)=\log_b(N)$, then $M=N$, but only when $M>0$ and $N>0$.

It also helps to keep the inverse relationship in mind visually. The exponential and logarithmic graphs are reflections across $y=x$.

Exponential equations

A few common forms show up over and over:

1. Same base

   $$8^{x-1}=4^{2x+1}$$

   Rewrite with base 2:

   $$2^{3x-3}=2^{4x+2}$$

   so $3x-3=4x+2$, giving $x=-5$.

2. Isolate and convert

   $$3^{2x+1}=5$$

   Convert:

   $$2x+1=\log_3 5$$

   so

   $$x=\frac{\log_3 5-1}{2}$$

   Keep this exact unless a decimal is requested.

3. Take $\ln$ of both sides

   $$2^{x+1}=5^{3-x}$$

   Then

   $$(x+1)\ln2=(3-x)\ln5$$

   and solve normally.

4. Substitute for repeated structure

   $$4^x-5(2^x)+4=0$$

   Let $u=2^x$. Then $4^x=(2^x)^2=u^2$, so

   $$u^2-5u+4=0$$

   giving $u=1$ or $u=4$, so $x=0$ or $x=2$.

Also remember $b^x=e^{x\ln b}$. That can make awkward bases easier to handle.

Logarithmic equations

• A single log often converts straight to exponential form:

  $$\log_3(2x-1)=2 \Rightarrow 2x-1=9 \Rightarrow x=5$$

• If you can combine logs, do it before solving:

  $$\ln(x-2)+\ln(x+1)=\ln10$$

  becomes

  $$\ln((x-2)(x+1))=\ln10$$

  so $(x-2)(x+1)=10$, then solve. The product rule $\ln A+\ln B=\ln(AB)$ needs $A>0$ and $B>0$, so here that means $x>2$.

If symbolic isolation gets ugly, graphs or numerical methods are fair game. For example, $2^x=x+2$ is usually solved by graphing intersections.

Restrictions and Extraneous Solutions

This is where a lot of quiz points disappear.

• Exponential restriction
  $b^u>0$ for every real $u$. So $b^u=0$ or $b^u=c$ with $c\le0$ has no real solution.

• Logarithmic restriction
  Every log argument must be strictly positive.

So in

$$\log_2(x-3)+\log_2(x+1)=4$$

you need $x-3>0$ and $x+1>0$, so $x>3$, before solving.

Extraneous solutions often come from:

• combining logs and then solving a polynomial,
• substitution that gives an impossible value like $u\le0$ when $u=2^x$,
• context limits such as time needing $x\ge0$.

Always check candidates in the original equation, or at least check every original log argument.

Solving Exponential and Logarithmic Inequalities

Monotonicity controls the direction.

• If $b>1$, $b^x$ and $\log_b x$ are increasing, so the inequality direction stays the same.
• If $0<b<1$, they are decreasing, so the direction reverses.

Exponential inequalities

$$\left(\frac13\right)^{2x-1}\ge9$$

Rewrite $9$ as $\left(\frac13\right)^{-2}$. Since the base is between 0 and 1, reverse the inequality:

$$2x-1\le-2$$

so $x\le-\frac12$.

Logarithmic inequalities

Domain first. Always.

$$\log_{1/2}(x-1)<-2$$

You need $x>1$. Rewrite $-2$ as $\log_{1/2}4$. Since base $1/2$ is decreasing:

$$x-1>4$$

so $x>5$.

Graphically, solving $F(x)>G(x)$ means finding where the graph of $F$ lies above $G$. Endpoints come from intersections and excluded domain values.

Undoing a transformed exponential or logarithm

The two forms are

$$f(x)=ab^{x-h}+k \qquad\text{and}\qquad f(x)=a\log_b(x-h)+k$$

Inverse of a transformed exponential

From $y=ab^{x-h}+k$, undo in reverse order: subtract $k$, divide by $a$, apply $\log_b$, add $h$.

$$f^{-1}(x)=h+\log_b\left(\frac{x-k}{a}\right)$$

Inverse of a transformed logarithm

From $y=a\log_b(x-h)+k$, undo by: subtract $k$, divide by $a$, convert to exponential form, add $h$.

$$f^{-1}(x)=h+b^{(x-k)/a}$$

These two graphs show the full pattern in action for one transformed exponential and one transformed logarithm.

Inverses of transformed exponential and logarithmic functions

Domain and range swap between a function and its inverse, and the graphs reflect across $y=x$. The labeled points make that swap easy to see. A quick check is composition: $f(f^{-1}(x))=x$ on the inverse’s domain.