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Introduction to 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)

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Problem 1 of 10

Problem 1 A combined reading

Evaluate $\log_5(5^{\sqrt2-1})+5^{\log_5 3}$ .
1. Hint 1
  An exponential and a logarithm with matching bases undo each other.
2. Hint 2
  In the first term the result is the exponent; in the second it is the positive argument.
Answer

$\sqrt2+2$ .

Full solution

Both logarithms have positive arguments.
The first term is $\sqrt2-1$ , and the second is $3$ .
$\log_5(5^{\sqrt2-1})+5^{\log_5 3}=\sqrt2+2$
The two inverse laws do not require the exponent to be an integer.

Answer

$\sqrt2+2$ .

Key idea

Cancellation depends on matching bases and valid arguments, not on whether the numbers are tidy.
Problem 1: Mark complete
Problem 2 A root inside

Evaluate $\log_{1/9}(\sqrt3)$ .
1. Hint 1
  The logarithm asks for the exponent that produces its argument.
2. Hint 2
  Write the base and the argument as powers of $3$ .
Answer

$-\frac14$ .

Full solution

Let the value be $y$ .
The definition gives
$(\frac19)^y=\sqrt3.$
In base $3$ this is
$3^{-2y}=3^{1/2}$
so $-2y=\frac12$ and
$y=-\frac14.$
Indeed
$(\frac19)^{-1/4}=\sqrt3.$

Answer

$-\frac14$ .

Key idea

A base between 0 and 1 needs a negative exponent to produce a value above one.
Problem 2: Mark complete
Problem 3 A reciprocal argument

Find the real domain of $h(x)=\log_4(\frac1{x+2})$ .
1. Hint 1
  The whole fraction must be defined and positive.
2. Hint 2
  Its numerator is positive, so its denominator must also be positive.
Answer

$x>-2$ .

Full solution

The denominator must be nonzero, and positivity of $1/(x+2)$ requires $x+2>0$ .
$x>-2$
At $x=-2$ the fraction is undefined; below $-2$ it is negative, so neither region is allowed.

Answer

$x>-2$ .

Key idea

A logarithm of a fraction requires a defined fraction with positive value.
Problem 3: Mark complete
Problem 4 A reflected curve

The graph shows $y=(\frac23)^x$ . Sketch its inverse on the same axes. State the inverse equation, its intercept on the horizontal axis, its domain, and its vertical asymptote.

The graph of y = (2/3)^x and the line y = x.

Text description of this figure

A grid with the x-axis and y-axis each from -3 to 4 on equal scales, a grid line and a label at every integer, and a short extra tick at two thirds on each axis. A solid decreasing curve, labeled y equals two thirds to the power x, enters at the left edge at a height of about 3.4, passes through the labeled points (-2, 9/4), (0, 1) and (1, two thirds), and flattens toward the x-axis without reaching it as x grows. A dashed straight line through the origin, labeled y equals x, rises from the lower left corner to the upper right corner. Nothing else is drawn.
1. Hint 1
  An inverse exchanges each point's input and output.
2. Hint 2
  Reflect the plotted curve across $y=x$ and exchange the roles of its domain and range.
Answer

$y=\log_{2/3}x$ ; intercept $(1,0)$ ; domain $x>0$ ; asymptote $x=0$ . The sketch is in the full solution.

Full solution

The inverse is the logarithm with base $\frac23$ .
The points $(0,1)$ , $(1,\frac23)$ , and $(-2,\frac94)$ exchange coordinates to give $(1,0)$ , $(\frac23,1)$ , and $(\frac94,-2)$ .

The reflected curve decreases for $x>0$ .
The horizontal asymptote $y=0$ becomes the vertical asymptote $x=0$ .
The exponential's positive range becomes the logarithm's domain.

The sketch of the inverse, y = log base 2/3 of x.

Answer

$y=\log_{2/3}x$ ; intercept $(1,0)$ ; domain $x>0$ ; asymptote $x=0$ . The sketch is in the full solution.

Key idea

Reflecting an exponential across the line y equals x exchanges its coordinates and its domain and range.
Problem 4: Mark complete
Problem 5 An output interval

A logarithm machine returns $y=\log_4 x$ . Find the positive inputs $x$ for which its output satisfies $-1\le y\le\frac12$ .
1. Hint 1
  The exponential with base $4$ preserves order.
2. Hint 2
  Convert the two endpoint outputs into inputs using powers of $4$ .
Answer

$\frac14\le x\le2$ .

Full solution

The logarithm and exponential are inverses, so $x=4^y$ .
Since $4^y$ increases, the extreme inputs occur at the extreme allowed outputs.
$4^{-1}=\frac14$
Also $4^{1/2}=2$ .
Thus the input interval is $\frac14\le x\le2$ , including both endpoints.

Answer

$\frac14\le x\le2$ .

Key idea

For $b>1$ , the relation $x=b^y$ carries $a\le y\le c$ exactly onto $b^a\le x\le b^c$ .
Problem 5: Mark complete
Problem 6 Two machine records

A machine follows $L(x)=\log_b x$ for a valid base $b$ . It records $L(u)=0$ and $L(v)=2$ , where $v=9u$ . Find $u$ , $v$ , and $b$ .
1. Hint 1
  An output of zero identifies the input through $b^0$ .
2. Hint 2
  Use the relation between the inputs before interpreting the output of two.
Answer

$u=1$ , $v=9$ , $b=3$ .

Full solution

The first record gives $u=b^0=1$ .
Then $v=9u=9$ .
The second record gives
$b^2=9$
and the positive base requirement selects $b=3$ .

The checks are $\log_3 1=0$ and $\log_3 9=2$ , so both records agree.

Answer

$u=1$ , $v=9$ , $b=3$ .

Key idea

A logarithm output identifies a power of the base even when the input is initially unknown.
Problem 6: Mark complete
Problem 7 Ordering without evaluating

Without a calculator, put $\log_{1/7}5$ , $\log_{1/7}\frac1{30}$ and $\log_{1/7}1$ in increasing order, and justify the order.
1. Hint 1
  Decide whether the function $\log_{1/7}x$ increases or decreases on $x>0$ .
2. Hint 2
  Order the three arguments, then reverse or keep that order.
Answer

$\log_{1/7}5<\log_{1/7}1<\log_{1/7}\frac1{30}$ .

Full solution

The logarithm with base $\frac17$ is the inverse of $(\frac17)^x$ , which decreases, so $\log_{1/7}x$ also decreases on $x>0$ : a larger positive argument gives a smaller logarithm.

The arguments satisfy $\frac1{30}<1<5$ , so their logarithms come in the reverse order.
As a check on the signs, $\log_{1/7}1=0$ , a power of $\frac17$ equal to $5$ needs a negative exponent, and one equal to $\frac1{30}$ needs a positive exponent.

Answer

$\log_{1/7}5<\log_{1/7}1<\log_{1/7}\frac1{30}$ .

Key idea

For $0<b<1$ , $\log_b x$ decreases on $x>0$ , so a larger positive argument gives a smaller logarithm.
Problem 7: Mark complete
Problem 8 An input claim

A student claims that $\log_x(13-x)$ is defined whenever $x<13$ . Determine its complete real domain and assess the claim.
1. Hint 1
  The variable plays two roles here, as the base and inside the argument.
2. Hint 2
  Check the conditions on the base and on the argument separately, then combine them.
Answer

No; the domain is $0<x<13$ with $x\ne1$ .

Full solution

A logarithm's base must be positive and not $1$ , so $x>0$ and $x\ne1$ .
Its argument must be positive, so $13-x>0$ , that is, $x<13$ .

All three conditions must hold at once, so the domain is $0<x<13$ with $x\ne1$ .
The claim fails: for example, $x=-2$ satisfies $x<13$ , but a negative base is not allowed, and at $x=1$ the base is $1$ .

Answer

No; the domain is $0<x<13$ with $x\ne1$ .

Key idea

When a variable is both the base and part of the argument, the domain must satisfy every condition on each.
Problem 8: Mark complete
Problem 9 An unrestricted claim

A student simplifies $5^{\log_5(t-1)}$ to $t-1$ and claims the equality holds for every real $t$ . Is that correct? State exactly where the simplification is valid.
1. Hint 1
  Cancellation does not create values where the original expression was undefined.
2. Hint 2
  Test the argument of the logarithm before applying the inverse law.
Answer

No; the equality is valid exactly when $t>1$ .

Full solution

The logarithm requires $t-1>0$ , which is $t>1$ .
On that domain the inverse law gives
$5^{\log_5(t-1)}=t-1$
At $t\le1$ , the original expression has no real value, even though the simplified expression does.

Answer

No; the equality is valid exactly when $t>1$ .

Key idea

An inverse-law simplification keeps the domain restriction of the original expression.
Problem 9: Mark complete
Problem 10 Where a function meets its inverse

For $f(x)=(\frac1{16})^x$ , a student claims that the graphs of $f$ and $f^{-1}$ can meet only on the line $y=x$ . Test the point $P=(\frac14,\frac12)$ on both graphs and assess the claim.
1. Hint 1
  Test membership in each graph separately.
2. Hint 2
  A point $(a,b)$ lies on the graph of $f^{-1}$ exactly when $(b,a)$ lies on the graph of $f$ .
Answer

$P$ lies on both graphs; the claim is false.

Full solution

On the graph of $f$ : $(\frac1{16})^{1/4}=\frac12$ , so $P$ lies on it.

On the graph of $f^{-1}$ : $P$ lies on it exactly when $(\frac12,\frac14)$ lies on the graph of $f$ , and $(\frac1{16})^{1/2}=\frac14$ , so it does.

The coordinates of $P$ differ, so $P$ is not on $y=x$ .
The two graphs meet off the diagonal, so the claim is false.

Answer

$P$ lies on both graphs; the claim is false.

Key idea

A function and its inverse can meet at points off the line y equals x, as (1/16)^x and its logarithm do.
Problem 10: Mark complete

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