Exponential and Logarithmic Equations and Inequalities

Solve $3\cdot 2^x+5=53$. Peel off what sits outside the power first. Subtract 5 to get $3\cdot 2^x=48$, then divide by 3 to get $2^x=16$. Now the power is alone, and $16=2^4$, so $x=4$.

When the number is not a power of the base, the logarithm is the answer. $2^x=20$ gives $x={\log }_{2}20=\frac{\ln 20}{\ln 2}$, about 4.32. Maya's account works the same way: $1000(1.06{)}^t=1500$ gives ${1.06}^t=1.5$, so $t=\frac{\ln 1.5}{\ln 1.06}$, about 6.96 years.

A logarithm comes off the other way, by turning it into a power. ${\log }_3(x-2)=4$ means $x-2=3^4=81$, so $x=83$.

When two powers have different bases, rewrite one so the bases match. Any $b^x$ can be written as $c^{({\log }_{c}b)x}$. Since ${\log }_{3}9=2$, the equation $9^x=3^{x+4}$ becomes $3^{2x}=3^{x+4}$. Equal powers of 3 have equal exponents, so $2x=x+4$ and $x=4$.

Solve ${\log }_{2}x+{\log }_2(x-3)=2$. Combine the logarithms into ${\log }_2(x(x-3))=2$, then turn it into a power: $x^2-3x=4$. That factors as $(x-4)(x+1)=0$, so the algebra hands you $x=4$ and $x=-1$.

Put each one back into the original equation. For $x=4$: ${\log }_{2}4+{\log }_{2}1=2+0=2$. It works. For $x=-1$: ${\log }_2(-1)$ has no value, because the input of a logarithm must be positive. So $x=-1$ is an extraneous solution. Your algebra produced it, but it does not solve the equation you started with. It slipped in because ${\log }_2(x(x-3))$ accepts $x=-1$ and the original two logarithms do not.

A negative answer is not automatically extraneous. In ${\log }_5(x+9)=1$, the answer $x=-4$ makes the input 5, which is fine.

A context can rule a value out too. If $t$ counts minutes after a tank starts to drain and the model holds only for $0\le t\le 6$, then $t=7$ is not an answer, even if it solves the equation. That goes for a value you read off a graph as well.

Solve ${0.5}^x>\frac{1}{8}$. Since $\frac{1}{8}={0.5}^3$, you want ${0.5}^x>{0.5}^3$. But ${0.5}^x$ is decreasing: bigger inputs give smaller outputs. So the outputs beat ${0.5}^3$ when the inputs are smaller than 3, and the solution is $x<3$. Try $x=4$: ${0.5}^4=\frac{1}{16}$, which is less than $\frac{1}{8}$.

With logarithms, the flip shows up in a particular step. Take $\ln$ of both sides: $x\ln 0.5>\ln \frac{1}{8}$. Taking $\ln$ keeps the direction, because $\ln$ is increasing. (A logarithm whose base is between 0 and 1, such as ${\log }_{0.5}$, is decreasing, so applying it reverses the inequality at once.) Now divide by $\ln 0.5$. Since 0.5 is between 0 and 1, $\ln 0.5$ is negative, and dividing by a negative number reverses the inequality: $x<\frac{\ln (1/8)}{\ln 0.5}=3$.

A logarithmic inequality needs its domain as well. For ${\log }_3(x-1)<2$, rewrite it as $x-1<3^2=9$, which keeps the direction because $3^x$ is increasing. The input must also be positive, so $x-1>0$. Together, $1<x<10$. Stopping at $x<10$ would let in $x=0$, where ${\log }_3(-1)$ has no value.

Take $f(x)=3\cdot 2^{x-1}+4$. Read what $f$ does to an input, in the order it happens: subtract 1, raise 2 to that power, multiply by 3, add 4.

The inverse runs that list backward, undoing each step. The last thing $f$ did is the first thing to undo. So subtract 4, divide by 3, take ${\log }_2$, and add 1:

Check it on a point. $f(3)=3\cdot 2^2+4=16$, and $f^{-1}(16)={\log }_{2}4+1=3$. You are back where you started.

You may have learned to swap $x$ and $y$ in $y=3\cdot 2^{x-1}+4$ and solve for $y$. That does exactly the same four steps in the same order. If the input is scaled first, as in $2^{x/10}$, the scaling is one more step on the list: divide by 10 going forward, so multiply by 10 last going back.

A function of the form $a{\log }_b(x+h)+k$ inverts the same way, and its inverse is exponential, because a power of $b$ is what undoes ${\log }_b$. Domain comes along for free: every output of $f$ is more than 4, and $f^{-1}$ accepts only $x>4$.