Why do logarithmic equations so often produce extraneous solutions?

Because a logarithm is only defined for a positive argument, so the original equation carries a domain restriction that the algebra takes no account of. The algebra can therefore hand you values that solve the rearranged equation but not the one you were given.

True or False? Two logarithms can be combined with the product or quotient property only if they have the same base.

True. Both properties are stated for a single base, so they do not apply across two different ones. Where the bases differ, use the change of base property first to bring them into line.

What is the domain of the inverse of a transformed logarithmic function, and why?

All real numbers. The domain of an inverse is the range of the original, and a logarithmic function in general form has range all real numbers. That fits the inverse being an exponential function, which is defined for every real input.

AP®MathsCollege BoardPrecalculusFlashcardsExponential & Logarithmic FunctionsExponential & Logarithmic Equations & Inequalities

Exponential & Logarithmic Equations & Inequalities (College Board AP® Precalculus): Flashcards

1/24

0Still learning

Know0

FrontSolving Exponential Equations & Inequalities
Solve $2^{(x + 1)} = 16$ by rewriting with a common base.

Was this flashcard helpful?

Cards in this collection (24)

Solve $2^{(x + 1)} = 16$ by rewriting with a common base.
$x = 3$ .
Write $16$ as $2^{4}$ , so the equation becomes $2^{(x + 1)} = 2^{4}$ .
With the same base on both sides the exponents must be equal, giving $x + 1 = 4$ .
Solve $e^{(x + 3)} = 10$ .
$x = \ln 10 - 3$ .
Take the natural logarithm of both sides, giving $\ln (e^{(x + 3)}) = \ln 10$ .
Since $\ln$ and $e$ are inverses they cancel on the left, leaving $x + 3 = \ln 10$ .
Fill in the two gaps.
To solve an equation such as $a e^{(b x + c)} + d = k$ , first $______$ the exponential term on one side, then take the $______$ of both sides.
The completed sentence is:
To solve an equation such as $a e^{(b x + c)} + d = k$ , first isolate the exponential term on one side, then take the logarithm of both sides.
Subtracting $d$ and dividing by $a$ does the isolating, and taking $\ln$ then brings the exponent down.
Solve $e^{2 x} + 2 e^{x} - 15 = 0$ .
$x = \ln 3$ .
Since $e^{2 x} = {(e^{x})}^{2}$ , substituting $y = e^{x}$ gives $y^{2} + 2 y - 15 = 0$ , which factors as $(y + 5) (y - 3) = 0$ .
The value $y = - 5$ has to be rejected because $e^{x}$ is always positive, leaving $e^{x} = 3$ .
True or False?
The inequality $0 . 5^{x} > 0.125$ has solution $x > 3$ .
False.
Solving the corresponding equation does give $x = 3$ , but $0.5 < 1$ , so $0 . 5^{x}$ is a decreasing function.
That reverses which side works, and the solution is $x < 3$ .
How do you solve an exponential inequality such as $e^{2 x} > 7$ ?
Solve the corresponding equation first, giving $2 x = \ln 7$ and so $x = \frac{\ln 7}{2}$ .
Then use the direction of the function to choose the side: here $e > 1$ makes $e^{2 x}$ increasing, so the solution is $x > \frac{\ln 7}{2}$ .
Why does $e^{x} = - 3$ have no solution?
Because $b^{x} > 0$ for every real $x$ when the base $b$ is positive.
An exponential expression can never produce a negative output, so no value of $x$ satisfies the equation.
True or False?
When solving an exponential equation you may take a logarithm to any base, not only base $e$ .
True.
Any base works, provided you apply the same one to both sides.
The natural logarithm is simply the most convenient choice when the equation is itself written with base $e$ .
Solve $\log_{5} x = 3$ .
$x = 125$ .
Convert straight to exponential form using the definition of a logarithm, which gives $x = 5^{3}$ .
If $\log_{b} A = \log_{b} B$ , what can you conclude, and what must you check?
You can conclude that $A = B$ , because a logarithmic function is invertible and so never produces the same output twice.
You must check that $A$ and $B$ are both positive, since a logarithm is only defined for a positive argument.
Solve $\log_{10} (x - 1) + \log_{10} (x + 3) = \log_{10} (x + 9)$ .
$x = 3$ .
The product property gives $(x - 1) (x + 3) = x + 9$ , which rearranges to $x^{2} + x - 12 = 0$ and factors as $(x + 4) (x - 3) = 0$ .
The value $x = - 4$ has to be rejected, because it makes two of the arguments negative.
Why do logarithmic equations so often produce extraneous solutions?
Because a logarithm is only defined for a positive argument, so the original equation carries a domain restriction that the algebra takes no account of.
The algebra can therefore hand you values that solve the rearranged equation but not the one you were given.
Dividing by $2$ solves $2 \log_{3} x = 4$ as $x = 9$ , but using the power property first gives $x = 9$ or $x = - 9$ . Which is correct?
Only $x = 9$ .
The two routes produce equations with different domains, since $\log_{3} x$ needs $x > 0$ while $\log_{3} (x^{2})$ is defined for every $x \neq 0$ .
A solution must satisfy the equation as it was given, and that equation contains $\log_{3} x$ .
True or False?
Two logarithms can be combined with the product or quotient property only if they have the same base.
True.
Both properties are stated for a single base, so they do not apply across two different ones.
Where the bases differ, use the change of base property first to bring them into line.
How do you solve $\ln (x - 1) > 3$ ?
Solve the corresponding equation first, giving $x - 1 = e^{3}$ .
Since $e > 1$ the natural logarithm is increasing, so the inequality holds for $x - 1 > e^{3}$ , that is $x > 1 + e^{3}$ .
Check the argument stays positive throughout, which it does here.
Fill in the two gaps.
After solving a logarithmic equation, substitute each answer back and reject it if any argument turns out to be $______$ or $______$ instead of positive.
The completed sentence is:
After solving a logarithmic equation, substitute each answer back and reject it if any argument turns out to be zero or negative instead of positive.
Every logarithmic expression in the original equation has to be checked, not just one of them.
How do you find the inverse of a transformed exponential function such as $f (x) = a b^{(x + h)} + k$ ?
Think of it as a sequence of operations applied to $x$ , then reverse each one in the opposite order.
Subtract $k$ , divide by $a$ , take $\log_{b}$ of both sides to undo the exponential, then subtract $h$ .
Find the inverse of $f (x) = 3 \cdot 2^{(x + 1)} - 5$ .
$f^{- 1} (x) = \log_{2} (\frac{x + 5}{3}) - 1$ .
Add $5$ and divide by $3$ , then take $\log_{2}$ , which undoes the power of $2$ and leaves $x + 1$ .
Subtracting $1$ and swapping the variables completes it.
Fill in the two gaps.
The inverse of a transformed exponential function is always a transformed $______$ function, and the inverse of a transformed logarithmic function is always a transformed $______$ function.
The completed sentence is:
The inverse of a transformed exponential function is always a transformed logarithmic function, and the inverse of a transformed logarithmic function is always a transformed exponential function.
That is what you would expect, given that the two base function types are inverses of each other.
For $f (x) = a b^{(x + h)} + k$ with $a > 0$ , what is the domain of $f^{- 1}$ ?
The domain of $f^{- 1}$ is $(k, \infty)$ .
That is the range of $f$ , because $a b^{(x + h)}$ is always positive and $k$ shifts it upwards.
It also matches the inverse itself, whose logarithm needs a positive argument.
Find the inverse of $f (x) = 2 \log_{3} (x - 4) + 1$ .
$f^{- 1} (x) = 3^{\frac{x - 1}{2}} + 4$ .
Subtract $1$ and divide by $2$ to reach $\frac{y - 1}{2} = \log_{3} (x - 4)$ .
Converting that to exponential form gives $3^{\frac{y - 1}{2}} = x - 4$ , so add $4$ and swap the variables.
What is the domain of the inverse of a transformed logarithmic function, and why?
All real numbers.
The domain of an inverse is the range of the original, and a logarithmic function in general form has range all real numbers.
That fits the inverse being an exponential function, which is defined for every real input.
Find $f^{- 1}$ for $f (x) = 5 e^{(x - 2)} + 3$ , and state its domain.
$f^{- 1} (x) = \ln (\frac{x - 3}{5}) + 2$ , with domain $x > 3$ .
Subtract $3$ , divide by $5$ , take $\ln$ to undo the $e$ , then add $2$ and swap the variables.
The domain follows from requiring $\frac{x - 3}{5} > 0$ .
True or False?
Checking that $f^{- 1} (f (x)) = x$ for one value of $x$ proves the inverse is correct.
False.
A single value is a quick way to catch an error, and it is well worth doing.
It cannot confirm the inverse, though, since a wrong expression can still happen to give the right output at one particular input.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Logarithmic FunctionsNext:Modeling with Exponential & Logarithmic Functions