Core practice ← Back to lesson

Properties of Logarithms: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A power of the base

For real $t$ and $y>0$ , expand $\log_{11}(11^t y)$ and simplify.
1. Hint 1
  Both factors are positive, so the product rule applies.
2. Hint 2
  The logarithm of a power of its own base is that exponent.
Answer

$t+\log_{11}y$ .

Full solution

The factors $11^t$ and $y$ are both positive, so the product rule splits the logarithm into $\log_{11}(11^t)$ plus $\log_{11}y$ .
Since $\log_{11}(11^t)=t$ for every real $t$ , the expansion is $t+\log_{11}y$ .

Answer

$t+\log_{11}y$ .

Key idea

The product rule splits off a power of the base, whose logarithm is its exponent.
Problem 1: Mark complete
Problem 2 A single argument

For $t>0$ , condense $\log_{13}(t^2+5)-\frac32\log_{13}t$ into one logarithm with coefficient one.
1. Hint 1
  Move the outside coefficient into an exponent before combining the logarithms.
2. Hint 2
  Subtraction places the second positive expression in the denominator.
Answer

$\log_{13}\left(\frac{t^2+5}{t^{3/2}}\right)$ , or equivalently $\log_{13}\left(\frac{t^2+5}{t\sqrt t}\right)$ .

Full solution

For $t>0$ both arguments are positive and $t^{3/2}$ is not zero.
The power rule turns $\frac32\log_{13}t$ into $\log_{13}(t^{3/2})$ , and the quotient rule gives
$\log_{13}\left(\frac{t^2+5}{t^{3/2}}\right)$
Expanding this recovers the original coefficients.

Answer

$\log_{13}\left(\frac{t^2+5}{t^{3/2}}\right)$ , or equivalently $\log_{13}\left(\frac{t^2+5}{t\sqrt t}\right)$ .

Key idea

Power and quotient rules combine logarithms while retaining their original positive-argument conditions.
Problem 2: Mark complete
Problem 3 Two bases

Evaluate $\frac{\log_2 7}{\log_4 7}$ .
1. Hint 1
  Write both logarithms in one common base.
2. Hint 2
  Base $4$ is $2^2$ , and $\log_2 7$ is nonzero.
Answer

$2$ .

Full solution

Change the denominator to base $2$ .
$\log_4 7=\frac{\log_2 7}{2}$
The numerator divided by half itself is $2$ .
Cancellation is legal since $7\ne1$ , so
$\log_2 7\ne0.$

Answer

$2$ .

Key idea

Changing both logarithms to one base can expose a common nonzero factor.
Problem 3: Mark complete
Problem 4 A reconstructed quantity

Positive numbers $U,V$ satisfy $\log_{10}U=2.4$ and $\log_{10}V=0.8$ . Find both $\log_{10}((U/V)^2)$ and $(\log_{10}(U/V))^2$ .
1. Hint 1
  The placement of the square determines whether the power rule applies.
2. Hint 2
  First find $\log_{10}(U/V)$ by subtracting the given values.
Answer

$\log_{10}((U/V)^2)=3.2$ ; $(\log_{10}(U/V))^2=2.56$ .

Full solution

The quotient is positive and $V\ne0$ .
Its logarithm is
$2.4-0.8=1.6.$
Squaring the argument doubles this logarithm, giving
$\log_{10}((U/V)^2)=3.2$

Squaring the logarithm instead gives
$(1.6)^2=2.56.$
The written parentheses decide which operation occurs first.

Answer

$\log_{10}((U/V)^2)=3.2$ ; $(\log_{10}(U/V))^2=2.56$ .

Key idea

Squaring a logarithm and taking the logarithm of a square are different operations.
Problem 4: Mark complete
Problem 5 A squared numerator

Expand $\log_5(\frac{(x+2)^2}{x^2+1})$ into logarithms with no powers on their arguments, preserving its full real domain. State that domain.
1. Hint 1
  The numerator is positive except where its squared factor is zero.
2. Hint 2
  Use an absolute value when moving the square outside the logarithm.
Answer

$2\log_5|x+2|-\log_5(x^2+1)$ ; $x\ne-2$ .

Full solution

The denominator $x^2+1$ is positive for every real $x$ , and the numerator is positive exactly when $x\ne-2$ .
Thus that is the full domain.

On it, $|x+2|>0$ .
The quotient and power rules give
$\begin{aligned}&\log_5((x+2)^2)\\&=2\log_5|x+2|\end{aligned}$
Subtracting $\log_5(x^2+1)$ completes the expansion without removing any allowed negative input.

Answer

$2\log_5|x+2|-\log_5(x^2+1)$ ; $x\ne-2$ .

Key idea

Absolute values preserve the full domain when expanding logarithms of even powers.
Problem 5: Mark complete
Problem 6 Bounds that decide

Using only $1.23<\log_{10}17<1.24$ , can you determine the number of digits of $17^{13}$ ? Explain. Then use $1.2304<\log_{10}17<1.2305$ to determine it.
1. Hint 1
  If both bounds on the logarithm lie strictly between the same two consecutive integers, the digit count is decided.
2. Hint 2
  Multiply both bounds by $13$ and see whether the interval crosses an integer.
Answer

No for the first bounds; the sharper bounds show $17^{13}$ has $16$ digits.

Full solution

By the power rule,
$\log_{10}(17^{13})=13\log_{10}17.$
The first bounds give $13\log_{10}17>15.99$ and
$13\log_{10}17<16.12.$
This interval contains $16$ , so the number could lie below $10^{16}$ (with $16$ digits) or at least $10^{16}$ (with $17$ ), and the bounds do not decide.

The sharper bounds give $13\log_{10}17>15.9952$ and
$13\log_{10}17<15.9965.$
Now $15<\log_{10}(17^{13})<16$ , so $10^{15}<17^{13}<10^{16}$ , and $17^{13}$ has $16$ digits.

Answer

No for the first bounds; the sharper bounds show $17^{13}$ has $16$ digits.

Key idea

For a positive integer $N$ , bounds that place $\log_{10}N$ in $d-1\le\log_{10}N<d$ show that $N$ has $d$ digits.
Problem 6: Mark complete
Problem 7 Three recorded logarithms

Let $p=\log_6 5$ and $q=\log_6 7$ . Write $\log_{36}(\frac{25}{7})$ in terms of $p,q$ .
1. Hint 1
  First convert base $36$ to base $6$ .
2. Hint 2
  The numerator involves a quotient and a square; the denominator is $\log_6 36$ .
Answer

$p-\frac q2$ .

Full solution

Change of base gives
$\begin{aligned}&\log_{36}(25/7)\\&=\frac{\log_6(25/7)}2\end{aligned}$
The denominator is $2$ , hence nonzero.

The numerator is $2\log_6 5-\log_6 7$ , or $2p-q$ .
Dividing by $2$ gives $p-\frac q2$ .
All numerical arguments are positive.

Answer

$p-\frac q2$ .

Key idea

Change of base and the logarithm rules can separate the effects of the base and the argument.
Problem 7: Mark complete
Problem 8 A rule from powers

Let $M=4^s$ and $N=4^t$ for real $s,t$ . A student claims $\log_4(M^2N)=2s+t$ . Is the claim correct? Explain directly with exponent laws. Also evaluate $\log_4(M/N)$ directly with exponent laws.
1. Hint 1
  Replace $M,N$ with their stated powers before taking a logarithm.
2. Hint 2
  Multiplying powers of the same base adds their exponents.
Answer

Yes; $\log_4(M^2N)=2s+t$ . Also $\log_4(M/N)=s-t$ .

Full solution

The powers $M,N$ are positive.
Squaring $M$ and multiplying by $N$ gives
$M^2N=4^{2s+t}$
The exponent that produces this argument is $2s+t$ , so its base-4 logarithm is $2s+t$ .
This proves the claim for every real $s,t$ .

Dividing the powers instead subtracts their exponents:
$\frac MN=4^{s-t}$
so
$\log_4(M/N)=s-t.$

Answer

Yes; $\log_4(M^2N)=2s+t$ . Also $\log_4(M/N)=s-t$ .

Key idea

Each logarithm rule is an exponent law read through the inverse operation.
Problem 8: Mark complete
Problem 9 A proposed simplification

A student replaces $\log_2(3x+1)$ by $\log_2 3+\log_2 x$ for $x>0$ . Is the replacement valid for any positive $x$ ? Explain.
1. Hint 1
  Combine the proposed replacement into one logarithm.
2. Hint 2
  Two logarithms of the same valid base agree only when their positive arguments agree.
Answer

No positive value of $x$ makes the replacement valid.

Full solution

The proposed right side combines to $\log_2(3x)$ .
Equality would require
$3x+1=3x$
which is impossible.
Both sides are defined for $x>0$ , so the failure is not a domain issue; the arguments differ.

Answer

No positive value of $x$ makes the replacement valid.

Key idea

A logarithm of a sum is not converted into a sum of logarithms by the product rule.
Problem 9: Mark complete
Problem 10 A new base

Let $b>0$ , $b\ne1$ , and $M>0$ . For which real $r$ is $\log_{b^r}M=\frac{\log_bM}r$ valid? Justify both the base restriction and the formula.
1. Hint 1
  First decide for which $r$ the number $b^r$ is a legal base.
2. Hint 2
  Then change the base of the left side to $b$ .
Answer

Exactly for $r\ne0$ .

Full solution

A base must be positive and not $1$ .
Since $b>0$ , $b^r>0$ for every $r$ ; and $b^r=1$ exactly when $r=0$ , because $b\ne1$ .
So $\log_{b^r}M$ is defined exactly when $r\ne0$ .

For $r\ne0$ , change of base gives
$\log_{b^r}M=\frac{\log_bM}{\log_b(b^r)}$
and $\log_b(b^r)=r$ , which is not zero, so the formula holds.
At $r=0$ neither side is defined.

Answer

Exactly for $r\ne0$ .

Key idea

Raising a base to a nonzero power divides its logarithms by that power.
Problem 10: Mark complete

← Back to lesson