Solving Exponential and Logarithmic Equations: Free Response

Both $125$ and $625$ are powers of $5$: $125=5^{3}$ and $625=5^{4}$. Rewrite each side on that base:

$5^{3(x+1)} = 5^{4(x-2)}.$

The two sides now share one base, so the one-to-one property lets you equate the exponents:

$3(x+1) = 4(x-2) \;\Longrightarrow\; 3x+3 = 4x-8 \;\Longrightarrow\; x = 11.$

Check it: $125^{12}=5^{36}$ and $625^{9}=5^{36}$, so both sides really do agree.

Both sides are powers of $3$: $27=3^{3}$ and $\dfrac19=3^{-2}$. Rewrite each side:

$3^{3x} = 3^{-2(x-4)}.$

Equate the exponents and solve:

$3x = -2(x-4) = -2x+8 \;\Longrightarrow\; 5x = 8 \;\Longrightarrow\; x = \frac{8}{5}.$

A fractional answer is not a sign of a mistake here; the exponents simply did not come out even.

Look back at what made parts A and B work. In each case, both sides could be rewritten as powers of a single base ($5$ in part A, $3$ in part B), and it was only after that rewriting that the one-to-one theorem applied at all:

$b^{u}=b^{v} \iff u=v \quad \text{for one fixed base } b>0,\ b\neq1.$

For $2^{x}=5^{x+1}$, no integer base $b$ lets both $2$ and $5$ be written as whole-number powers of $b$: both are prime, so neither can be written as a whole-number power of the other, and no smaller integer serves as a common base for both either. The technique from A and B has nothing to grab onto.

The equation is not stuck, though, because a different one-to-one property is available. $2^{x}$ and $5^{x+1}$ are both positive for every real $x$, and the logarithm is itself one-to-one on positive numbers: $\log_b M=\log_b N$ exactly when $M=N$, for $M,N>0$. That is what licenses applying $\log$ to both sides of the equation, even though the two sides were built from different bases.

No integer power connects $3$ and $50$, so take the common logarithm of both sides, which is legal since both sides are positive:

$\log\!\left(3^{x}\right) = \log 50 \;\Longrightarrow\; x\log 3 = \log 50 \;\Longrightarrow\; x = \frac{\log 50}{\log 3} = \log_3 50.$

Substituting the given approximations,

$x \approx \frac{1.6990}{0.4771} \approx 3.561.$

The power is buried under a coefficient, so isolate it first by dividing both sides by $6$:

$4^{x+2} = 150.$

Now take the common logarithm of both sides and use the power rule:

$(x+2)\log 4 = \log 150 \;\Longrightarrow\; x+2 = \frac{\log 150}{\log 4} \;\Longrightarrow\; x = \log_4 150 - 2.$

Substituting the given approximations,

$x \approx \frac{2.1761}{0.6021} - 2 \approx 3.614 - 2 = 1.614.$

An extraneous root can appear whenever a step widens the set of $x$ for which an equation makes sense, because then a value can satisfy the new, wider equation without satisfying the original narrower one.

Check whether that can happen here. Before the logarithm is taken, $b^{x}=k$ already makes sense at every real $x$: the left side $b^{x}$ is a positive number for every real $x$, and $k$ is given as a positive number too, so nothing about the equation restricts $x$ at all.

Taking the logarithm changes $b^{x}=k$ into $\log(b^{x})=\log k$, and $\log$ is defined at every input encountered here, since both $b^{x}$ and $k$ are positive throughout:

$b^{x}=k \;\Longrightarrow\; x\log b = \log k.$

So the new equation makes sense at exactly the same set of $x$ as the old one: no widening occurred. And because $\log$ is one-to-one, $\log(b^{x})=\log k$ holds for exactly the same $x$ as $b^{x}=k$ did, not merely for a superset of them.

Contrast this with combining two logarithms, where the ORIGINAL equation is defined only when each individual argument is positive, while the combined equation only needs their product to be positive: that is a genuine widening, which is why extraneous roots turn up there and not here.

Each logarithm in the equation needs its own argument to be positive, independently of the other. $\log_3(x+6)$ needs $x+6>0$, that is $x>-6$. $\log_3 x$ needs $x>0$.

Both conditions have to hold at once for the equation to be defined at all, so the domain is their intersection:

$x > -6 \ \text{ and } \ x>0 \;\Longrightarrow\; x>0.$

The stricter of the two conditions, $x>0$, is the one that actually decides the domain.

Combine with the product rule, then convert straight to exponential form:

$\log_3\!\big(x(x+6)\big) = 3 \;\Longrightarrow\; x(x+6) = 3^{3} = 27.$

That is a quadratic. Expand and factor:

$x^{2}+6x-27 = 0 \;\Longrightarrow\; (x-3)(x+9) = 0,$

so the candidates are $x=3$ and $x=-9$.

Test each candidate in the ORIGINAL equation, not the combined one, since the combined equation is what needs checking against.

At $x=3$: the arguments are $x+6=9$ and $x=3$, both positive, so the original equation is defined and

$\log_3 9 + \log_3 3 = 2+1 = 3,$

which matches. So $x=3$ is a genuine solution, and it also satisfies the domain $x>0$ found in part A.

At $x=-9$: the arguments are $x+6=-3$ and $x=-9$, both negative. $\log_3(-3)$ and $\log_3(-9)$ do not exist, so the original equation is not even defined at $x=-9$, and it fails the domain $x>0$ immediately.

It is worth seeing exactly why $x=-9$ showed up in part B anyway. The combined equation only asked for the PRODUCT $x(x+6)$ to equal $27$, and a product of two negative numbers is positive: $(-3)(-9)=27$ is exactly $27$, so the combined equation is perfectly satisfied there. Combining the two logarithms replaced a condition needing EACH argument positive with a weaker condition needing only their product positive, and $x=-9$ lives precisely in the gap between those two conditions.

Since $9=3^{2}$, the first term rewrites in terms of the same power as the rest of the equation:

$9^{x} = \left(3^{2}\right)^{x} = 3^{2x} = \left(3^{x}\right)^{2}.$

Substitute $u=3^{x}$. The equation becomes an ordinary quadratic in $u$:

$u^{2} - 6u - 27 = 0 \;\Longrightarrow\; (u-9)(u+3) = 0 \;\Longrightarrow\; u = 9 \ \text{ or } \ u=-3.$

Undo the substitution one root of $u$ at a time.

From $u=9$: $3^{x}=9=3^{2}$, and since both sides now share the base $3$, the one-to-one property gives

$3^{x}=3^{2} \;\Longrightarrow\; x=2.$

From $u=-3$: $3^{x}=-3$. An exponential with a positive base is positive for every real input, so $3^{x}$ never reaches a negative number. There is no real $x$ satisfying this equation at all.

Check the surviving value in the original equation: $9^{2}-6\cdot3^{2}-27 = 81-54-27=0$. The solution is $x=2$.

The two kinds of failure happen at different stages of the work, and it is worth separating them clearly.

An extraneous solution, of the kind combining two logarithms can produce, is a real value of $x$ that satisfies the derived equation and is then rejected only after checking it against the original equation's domain. The candidate exists the whole time; it simply does not survive the check.

Here the situation is different. Undoing $u=-3$ means solving $3^{x}=-3$ for $x$. Since the range of $3^{x}$ is the positive numbers,

$3^{x} > 0 \quad \text{for every real } x,$

no real number $x$ ever makes $3^{x}$ equal to $-3$, or to any number that is zero or negative. So this equation has no solution at all, and there is no value of $x$ to write down, test, or reject: the failure is not that a candidate got rejected, it is that no candidate was ever produced.

The distinction matters because it changes what you write on paper. For an extraneous solution, you record the candidate and then explain why it fails. For a substitution value like $u=-3$, you record that undoing it is impossible and move on; there is nothing to test.

Substituting $u=\log_2 x$ directly turns every term into a term in $u$:

$u^{2} - 5u + 6 = 0 \;\Longrightarrow\; (u-2)(u-3) = 0 \;\Longrightarrow\; u=2 \ \text{ or } \ u=3.$

Undo $u=\log_2 x$ by converting each value of $u$ back to exponential form,

$x = 2^{u}.$

From $u=2$: $x=2^{2}=4$. From $u=3$: $x=2^{3}=8$.

Check both in the original equation. At $x=4$, $\log_2 4=2$, so $(2)^{2}-5(2)+6=4-10+6=0$. At $x=8$, $\log_2 8=3$, so $(3)^{2}-5(3)+6=9-15+6=0$. Both survive.

Compare the two undoing steps directly, since they run in opposite directions.

When the substitution is $u=b^{x}$, undoing a found value of $u$ means solving $b^{x}=u$ for $x$. That equation has a solution only when $u$ is in the RANGE of $b^{x}$, which is the positive numbers $(0,\infty)$; a value like $u=-3$ falls outside that range, so no $x$ exists.

When the substitution is $u=\log_b x$, undoing a found value of $u$ means solving $\log_b x=u$ for $x$, which converts directly to

$x = b^{u}.$

This time the question is whether $b^{u}$ is defined, and $b^{u}$ is defined and positive for EVERY real number $u$, with no restriction at all. Whatever real number the quadratic in $u$ hands back, $x=b^{u}$ produces a genuine positive value of $x$.

The asymmetry traces back to the domain and range swapping between a function and its inverse. $b^{x}$ has domain all reals and range only the positive numbers, so undoing FROM a value of $u$ back to $x$ (when $u$ came from $b^x$) can fail, because $u$ has to land in that restricted range. $\log_b x$ has domain only the positive numbers and range all reals, so undoing it runs the opposite way: $x=b^u$ asks $b$ to accept ANY real exponent, which it always does.