Exercises — Logarithm — definition as inverse of exponential

Ask the question the log is asking: "$5$ to what power gives $25$?" We know $5^2=25$, so the exponent is $2$. ${\log }_{5}25=2$

The definition says $b^x=y⟺{\log }_{b}y=x$. Here $b=3$, $x=4$, $y=81$. ${\log }_{3}81=4$ Same fact, two costumes.

"$10$ to what power gives $1000$?" Counting zeros: $10^3=1000$. ${\log }_{10}1000=3$

"$2$ to what power gives $\frac{1}{16}$?" A fraction below $1$ needs a negative exponent when the base is $>1$. Since $2^4=16$, we have $2^{-4}=\frac{1}{16}$. ${\log }_2\frac{1}{16}=-4$

"$9$ to what power gives $3$?" We need the half-step, because $3$ is the square root of $9$. $9^{1/2}=\sqrt{9}=3$, so ${\log }_{9}3=\frac{1}{2}$

"$7$ to what power gives $1$?" Any nonzero base to the power $0$ equals $1$: $7^0=1$. ${\log }_{7}1=0$ This is why every log graph passes through $(1,0)$.

The base is $\frac{1}{2}$, which is between 0 and 1, so raising it to a positive power makes things smaller, and to a negative power makes them bigger. Look at the falling yellow curve in the figure above.

• "$\frac{1}{2}$ to what power gives $8$?" We must grow, so the exponent is negative: ${\left(\frac{1}{2}\right)}^{-3}=2^3=8$. Hence ${\log }_{1/2}8=-3$
• "$\frac{1}{2}$ to what power gives $\frac{1}{8}$?" Now we shrink, needing a positive exponent: ${\left(\frac{1}{2}\right)}^3=\frac{1}{8}$. Hence ${\log }_{1/2}\frac{1}{8}=3$ Notice the signs flip compared with a base $>1$ — that is the whole personality of a $0<b<1$ logarithm.

These are the two golden identities $b^{{\log }_{b}y}=y$ and ${\log }_b(b^x)=x$ from the parent note.

• $6^{{\log }_{6}13}=13$ — raise $6$ to the exponent that by definition gives $13$.
• ${\log }_4(4^{-0.5})=-0.5$ — the exponent is written right there.

Why not just take a log immediately? Because both sides are powers of $2$, so we can equate exponents exactly using the equal-base rule from the toolkit ($a^u=a^v\Rightarrow u=v$, valid because $a^x$ is one-to-one). $8=2^3$ and $32=2^5$, so $8^x=(2^3{)}^x=2^{3x}=2^5⟹3x=5⟹x=\frac{5}{3}$ The step "$2^{3x}=2^5\Rightarrow 3x=5$" is exactly the equal-base rule in action.

This is log form. Convert to exponential form — that's the whole trick. ${\log }_{3}x=4⟺3^4=x⟹x=81$

The unknown is the base. Convert to exponential form: ${\log }_{x}27=3⟺x^3=27⟹x=3$ Check the base is legal: $x=3>0$ and $\ne 1$. ✓

Common base $5$: $25=5^2$ and $\frac{1}{5}=5^{-1}$. Then apply the equal-base rule: $5^{2x}=5^{-1}⟹2x=-1⟹x=-\frac{1}{2}$ Sanity check: base $>1$ giving a result below $1$ ⇒ exponent negative. ✓

Break $12$ into the pieces we know: $12=2^2\times 3$. Product rule → sum; power rule → the exponent comes down: ${\log }_{b}12={\log }_b(2^2\cdot 3)=2{\log }_{b}2+{\log }_{b}3=2(0.4)+0.6=1.4$

Quotient rule turns a difference of logs into a log of a quotient: ${\log }_{2}40-{\log }_{2}5={\log }_2\frac{40}{5}={\log }_{2}8=3$

Change-of-base formula (from the toolkit): for any legal base $a$, ${\log }_{b}N=\frac{{\log }_{a}N}{{\log }_{a}b}.$ Taking $a=2$, $b=4$, $N=64$: ${\log }_{4}64=\frac{{\log }_{2}64}{{\log }_{2}4}.$ ${\log }_{2}64=6$ (since $2^6=64$) and ${\log }_{2}4=2$, so ${\log }_{4}64=\frac{6}{2}=3.$ Check directly: $4^3=64$. ✓

Product rule collapses the left side into one log: ${\log }_2(x(x-2))=3⟺x(x-2)=2^3=8$ $x^2-2x-8=0⟹(x-4)(x+2)=0⟹x=4\text{ or }x=-2$ Now check the domain ($y>0$ needed inside each log): $x=-2$ makes ${\log }_2(-2)$ undefined — reject it. $x=4$ gives ${\log }_{2}4+{\log }_{2}2=2+1=3$. ✓ $\boxed{x=4}$

Step 1 — find $k$. On $y=2^x$ at $x=3$: $k=2^3=8$. Step 2 — find $m$. On $y={\log }_{2}x$ at $x=k=8$: $m={\log }_{2}8=3$. So $k=8$, $m=3$. Step 3 — the geometry. The exponential passes through $(3,8)$; the logarithm passes through $(8,3)$. The coordinates are swapped. That is exactly what reflection in the line $y=x$ does — see the figure below and Inverse functions and reflection in $y=x$.

Spot the hidden quadratic. Let $u=2^x$ (so $u>0$ always). Then $2^{2x}=(2^x{)}^2=u^2$: $u^2-5u+4=0⟹(u-1)(u-4)=0⟹u=1\text{ or }u=4$ Undo the substitution:

• $2^x=1⟹x={\log }_{2}1=0$
• $2^x=4⟹x={\log }_{2}4=2$ Both give $u>0$, so both are valid: $x=0\text{ or }x=2$

The bases differ ($3$ and $5$) with no common base — this is precisely when we take logs of both sides (Solving exponential equations). Which base of log? It does not matter — any base works, because taking a log is applying the same one-to-one function to both sides. Base $10$, base $e$, or base $2$ all give the same final $x$; the ratio of logs just gets written differently. We'll pick ${\log }_{10}$ because it's on every calculator. ${\log }_{10}\left(3^{x+1}\right)={\log }_{10}\left(5^x\right)$ Power rule brings exponents down (the whole exponent): $(x+1){\log }_{10}3=x{\log }_{10}5$ Expand and collect the $x$ terms: $x{\log }_{10}3+{\log }_{10}3=x{\log }_{10}5$ ${\log }_{10}3=x({\log }_{10}5-{\log }_{10}3)$ $x=\frac{{\log }_{10}3}{{\log }_{10}5-{\log }_{10}3}\approx \frac{0.4771}{0.6990-0.4771}\approx 2.151$ (Redo it with $\ln$ and you get the identical $2.151$ — proof the base choice was free.)

Proof. By the change-of-base formula into (any) common base, say $\ln$: ${\log }_{b}a=\frac{\ln a}{\ln b},{\log }_{a}b=\frac{\ln b}{\ln a}$ Multiply — the pieces cancel: ${\log }_{b}a\cdot {\log }_{a}b=\frac{\ln a}{\ln b}\cdot \frac{\ln b}{\ln a}=1$ Application. With $a=2,b=8$: ${\log }_{8}2=\frac{1}{3}$ (since $8^{1/3}=2$) and ${\log }_{2}8=3$. $\frac{1}{3}\cdot 3=1✓$