Rewriting Exponential Forms
You said it yourself today: not comfortable yet, need more reps. That's exactly what this page is — the same four moves from this morning, worked again slowly, before you try any of them cold. Nothing here is new teaching; it's reps.
The One Idea Behind All Four Types
Every one of these problems is the same move: a coefficient and a piece of the base's power are interchangeable. You're never adding or subtracting anything — only multiplying and dividing, moving a factor of the base between the coefficient out front and the exponent. "Missing" a factor in the exponent? Steal it from the coefficient. Coefficient has an extra factor of the base hiding in it? Push it into the exponent. Four types below, one underlying move.
Borrow It Back — Shifted Exponent
The exponent isn't a clean $x$ — it's $x$ shifted by a number. Rewrite the coefficient so it "gives back" or "borrows" exactly what the shift is missing.
Same move, other direction. $9\cdot5^{x+1}$ — now the exponent has an extra $5$ it doesn't need. Pull that $5$ out into the coefficient instead: $9\cdot5\cdot5^{x}=45\cdot5^{x}$. Shifted down → coefficient shrinks. Shifted up → coefficient grows. Same rule, opposite direction.
Your turn. $20\cdot5^{x-1}=\,?$
Hint ›
The exponent is missing one $5$. Split $20$ into a $5$ and something else.
Check it ›
$20=4\cdot5$. Give the $5$ back: $4\cdot5\cdot5^{x-1}=4\cdot5^{x}$.
One more. $24\cdot3^{x+1}=\,?$
Hint ›
This time the exponent has an extra $3$ — which direction does the factor move?
Check it ›
$3^{x+1}=3\cdot3^{x}$, so $24\cdot3\cdot3^{x}=72\cdot3^{x}$.
Fraction Coefficient + Shifted Exponent
Same borrow-back idea — except the "missing factor" shows up as a fraction, which is really a negative exponent in disguise.
Your turn. $\dfrac{3}{25}\cdot5^{x+2}=\,?$
Hint ›
$25=5^2$, so $\dfrac{3}{25}=3\cdot5^{-2}$. Combine that with the $5^{x+2}$ first.
Check it ›
$5^{-2}\cdot5^{x+2}=5^{x}$, so the whole thing is just $3\cdot5^{x}$ — the shift and the fraction cancel completely.
Roots Hiding in the Exponent
This is the one that felt "awful" today, so it gets three full worked examples before you touch a practice problem — no fading, no shortcuts yet. The rule, in your tutor's words (borrowed from his sister): "the top stays to the right, the bottom slides to the left." For a power raised to a fraction, the numerator stays as the exponent — the denominator becomes a root.
Not every answer comes out as a whole number in front — a fraction coefficient is a perfectly good final answer.
Your turn — one practice problem, hints available. $\dfrac{2}{9}\cdot27^{(x+2)/3}=\,?$
Hint 1 ›
Start with the root rule only: what does $27^{(x+2)/3}$ become?
Hint 2 ›
$27^{(x+2)/3}=3^{x+2}=9\cdot3^{x}$. Now bring in the $\dfrac{2}{9}$ coefficient.
Check it ›
$\dfrac{2}{9}\cdot9\cdot3^{x}=2\cdot3^{x}$.
Two Points, Solve for A and B
Cover the steps below and try this cold first — this is the exact problem from this morning, and you got further on it than any of the others without help.
Find $f(x)=A\cdot B^{x}$ through $(0,-3)$ and $(2,-27)$.
Check your attempt ›
Plug in $x=0$ first — always. $B^{0}=1$ no matter what $B$ is, so the first point hands you $A$ directly: $-3=A\cdot1\Rightarrow A=-3$.
Now use the second point. $-27=A\cdot B^{2}=-3\cdot B^{2}$. Divide: $B^{2}=9$.
Solve, and check both roots. $B=3$ or $B=-3$ — but a base has to stay positive, so $B=-3$ is rejected.
Spot the error. A student solved the exact same problem above and wrote: "$A=-3$. Then $B^2=9$, and since $A$ was negative, I took $B=-3$ to match. Final answer: $f(x)=-3\cdot(-3)^{x}$." What's wrong, and why does it matter that the sign came from $A$, not $B$?
What went wrong ›
A base can never be negative — try plugging in $x=\tfrac12$ into $(-3)^{x}$ and it stops making sense (square root of a negative number). The negative sign belongs entirely to $A$; $B$ is always the positive root. $B^2=9$ has two solutions algebraically, but only one is a legal base.
New numbers, same method. $f(x)=A\cdot B^{x}$ through $(0,4)$ and $(1,12)$.
Hint ›
What does plugging in $x=0$ give you immediately?
Check it ›
$A=4$. Then $12=4\cdot B\Rightarrow B=3$. $f(x)=4\cdot3^{x}$.
Cold Recall — All Four Types, Mixed Up
This is today's consolidation pass, not the long-term spaced review — expect some misses, and we'll redo this page with new numbers in a few days once these are more automatic.
1 · $8^{x/3}=\,?$ in the form $B^{x}$.
Check ›
$8=2^3$, so $8^{x/3}=2^{x}$.
2 · $\dfrac{1}{4}\cdot8^{x/3}=\,?$
Check ›
$8^{x/3}=2^{x}$ from question 1, so this is $\dfrac{1}{4}\cdot2^{x}$ — a fractional coefficient is a fine final answer.
3 · $18\cdot4^{x-1}=\,?$
Check ›
$18=4.5\cdot4$ — messy on purpose. Not every coefficient splits into whole numbers; here it's cleaner to write $4^{x-1}=\dfrac{4^{x}}{4}$, so $18\cdot\dfrac{4^{x}}{4}=4.5\cdot4^{x}$.
4 · $\dfrac{5}{49}\cdot7^{x+2}=\,?$
Check ›
$49=7^2$, so $\dfrac{5}{49}=5\cdot7^{-2}$, and $7^{-2}\cdot7^{x+2}=7^{x}$. Answer: $5\cdot7^{x}$.
5 · $f(x)=A\cdot B^{x}$ through $(0,2)$ and $(3,54)$.
Check ›
$A=2$. $54=2\cdot B^{3}\Rightarrow B^{3}=27\Rightarrow B=3$. $f(x)=2\cdot3^{x}$.
6 · In your own words: why does plugging in $x=0$ always find $A$ first, no matter what $B$ turns out to be?
Check ›
Any nonzero number raised to the $0$ power is $1$, so $B^{0}=1$ always — the equation collapses to $y=A\cdot1$, isolating $A$ before $B$ is even known.
7 · $\dfrac{3}{2}\cdot8^{(x-1)/3}=\,?$
Check ›
$8^{(x-1)/3}=2^{x-1}=\dfrac{2^{x}}{2}$. So $\dfrac{3}{2}\cdot\dfrac{2^{x}}{2}=\dfrac{3}{4}\cdot2^{x}$.
8 · A student rewrote $\dfrac{2}{9}\cdot27^{(x+2)/3}$ and got $6\cdot3^{x}$ instead of $2\cdot3^{x}$. What number did they mix up?
Check ›
They likely used $\dfrac{2}{3}$ instead of $\dfrac{2}{9}$ as the starting coefficient — that's the Example C problem from Type 3 above, not this one. Same method, different starting fraction, different answer. Always double-check which coefficient the problem actually gave you.
Before You Leave — Cheat-Sheet Line
One line: "Coefficient and base-power trade places — never add or subtract, only multiply/divide. Fractions are negative exponents. A root is a fraction in the exponent — top stays the exponent, bottom becomes the root. On two points, plug in x=0 first for A."