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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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra I. You can skip it.

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Problem 1 of 10

Problem 1 Seven to the power $x+2$

Find the exact value of $x$ satisfying $7^{x+2}=250$ , written using common logarithms.
1. Hint 1
  The unknown sits in an exponent, so look for the law that moves an exponent out to the front.
2. Hint 2
  Take the common logarithm of both sides, then apply the power law to $\log(7^{x+2})$ .
Answer

$x=\frac{\log(250)}{\log(7)}-2$ , equivalently $\frac{\log(250)-2\log(7)}{\log(7)}$ or $\frac{\log(250/49)}{\log(7)}$ .

Full solution

Both sides are positive, so their common logarithms are equal:
$\log(7^{x+2})=\log(250)$

The power law brings the whole exponent down as a coefficient:
$(x+2)\log(7)=\log(250)$

Since $7\ne1$ , the number $\log(7)$ is not zero, so dividing by it is allowed:
$x+2=\frac{\log(250)}{\log(7)}$

Subtracting $2$ gives
$x=\frac{\log(250)}{\log(7)}-2.$
Checking, this makes the exponent $x+2$ equal to $\log(250)/\log(7)$ , which is the power that turns $7$ into $250$ .

Answer

$x=\frac{\log(250)}{\log(7)}-2$ , equivalently $\frac{\log(250)-2\log(7)}{\log(7)}$ or $\frac{\log(250/49)}{\log(7)}$ .

Key idea

Taking a logarithm of both sides brings a whole exponent down as a coefficient, even when that exponent is a sum.
Problem 1: Mark complete
Problem 2 The matching displays

Find the positive number $u$ for which the common logarithm $\log(u)$ equals $\ln(e^{-3})$ .
1. Hint 1
  Common and natural logarithms use different bases, even when their displayed values agree.
2. Hint 2
  First undo the natural exponential, then ask which power of $10$ gives $u$ .
Answer

$u=0.001$ , or $10^{-3}$ .

Full solution

The natural logarithm undoes the base- $e$ power:
$\ln(e^{-3})=-3$

So $\log(u)=-3$ , which means
$u=10^{-3}$
Therefore $u=0.001$ , a positive number.
Its common logarithm and the given natural logarithm both equal $-3$ .

Answer

$u=0.001$ , or $10^{-3}$ .

Key idea

Equal logarithm outputs can correspond to different inputs when their bases differ.
Problem 2: Mark complete
Problem 3 One step up the logarithm

For positive numbers $U,V$ and a base $b>0$ with $b\ne1$ , the records say $\log_b(U)=q$ and $\log_b(V)=q+1$ . Find $V/U$ in terms of $b$ .
1. Hint 1
  An increase in the logarithm describes an increase in the exponent.
2. Hint 2
  Write $U$ and $V$ as powers of $b$ , then divide the positive powers.
Answer

$V/U=b$ .

Full solution

The definitions give $U=b^q$ and $V=b^{q+1}$ .
Since $U>0$ , their quotient is defined:
$\frac VU=\frac{b^{q+1}}{b^q}$

The exponent difference is $1$ , so
$\frac VU=b$
Multiplying $U$ by this factor produces $V$ , as the two logarithm records require.

Answer

$V/U=b$ .

Key idea

A unit increase in a logarithm multiplies its positive input by the base.
Problem 3: Mark complete
Problem 4 The highlighted curve

The figure shows $y=(\frac12)^x$ , with one portion highlighted. Sketch the portion of the inverse function, $y=\log_{1/2}(x)$ , corresponding to the highlighted portion, including its endpoints. Give the input interval of your sketch.

The curve $y=(\tfrac12)^x$ with one portion highlighted, and the dashed diagonal $y=x$ .

Text description of this figure

A square coordinate grid. The horizontal x-axis and the vertical y-axis each run from negative three to five, with gridlines, tick marks and number labels at every whole number and equal unit lengths on both axes, and an extra labeled tick at one quarter on each positive axis. A dashed straight line labeled y equals x runs from the bottom left corner through the origin to the top right corner. A light curve labeled y equals one half to the power x falls from the top of the window on the left, crosses the vertical axis at height one, and flattens out just above the x-axis on the right. The part of that curve between x equals negative two and x equals two is drawn as a thick highlighted line. Its two ends carry filled dots, labeled with the coordinate pairs (negative 2, 4) on the left and (2, one quarter) on the right, and a third filled dot with no label sits where the curve crosses the vertical axis at height one. Nothing else is drawn: there is no second curve, and no other points are marked.
1. Hint 1
  An inverse swaps the two coordinates of every point.
2. Hint 2
  Swap the coordinates of each labeled endpoint, then read off the smallest and largest input of the reflected piece.
Answer

Endpoints $(\frac14,2)$ and $(4,-2)$ ; sketched inputs $\frac14\le x\le4$ .

Full solution

The highlighted exponential runs from $(-2,4)$ to $(2,\tfrac14)$ .
Reflection swaps these endpoints to $(4,-2)$ and $(\tfrac14,2)$ , and swaps $(0,1)$ to $(1,0)$ .

The reflected portion belongs to
$y=\log_{1/2}(x)$
It falls as $x$ increases and includes inputs from $\tfrac14$ to $4$ , with both endpoints included.

Answer

Endpoints $(\frac14,2)$ and $(4,-2)$ ; sketched inputs $\frac14\le x\le4$ .

Key idea

Reflecting a restricted exponential portion swaps its endpoint coordinates and turns its output interval into the inverse input interval.
Problem 4: Mark complete
Problem 5 Readings six apart

Two positive readings $U$ and $V$ satisfy $\log_3(U)=2\log_3(V)$ and $U-V=6$ . Find both readings.
1. Hint 1
  The logarithm relationship can be translated into an ordinary relationship between the positive readings.
2. Hint 2
  Use the power law on $2\log_3(V)$ , then substitute into the difference equation.
Answer

$U=9$ , $V=3$ .

Full solution

Because $V>0$ , the power law gives
$2\log_3(V)=\log_3(V^2).$
The logarithm is one-to-one, so $U=V^2$ .
Substituting into the difference gives
$\begin{aligned}V^2-V&=6\\V^2-V-6&=0\\(V-3)(V+2)&=0\end{aligned}$

The candidates are $V=3$ and $V=-2$ , but positivity excludes $-2$ .
Thus $V=3$ and $U=9$ .
Checking, $\log_3(9)=2\log_3(3)=2$ and $9-3=6$ .

Answer

$U=9$ , $V=3$ .

Key idea

Logarithm laws can convert relationships between positive quantities into ordinary algebraic equations.
Problem 5: Mark complete
Problem 6 From $12$ to $50$

A positive quantity follows $Q(t)=Q_0e^{0.3t}$ for $t\ge0$ , where $Q_0>0$ and $t$ is in hours. It is observed at $Q=12$ and later at $Q=50$ . Find the exact elapsed time between those observations, using $\ln$ .
1. Hint 1
  Dividing the later quantity by the earlier one removes the unknown starting amount.
2. Hint 2
  The ratio is $e^{0.3\Delta t}$ , where $\Delta t$ is the elapsed time.
Answer

$\frac{\ln(25/6)}{0.3}$ hours, equivalently $\frac{10}{3}\ln(25/6)$ hours.

Full solution

Let the two observation times be $t_1<t_2$ .
Their positive ratio is
$\frac{50}{12}=e^{0.3(t_2-t_1)}$

Taking natural logarithms and using the power law gives
$0.3(t_2-t_1)=\ln(25/6)$

Dividing by $0.3\ne0$ gives the elapsed time $\ln(25/6)/0.3$ hours.
Substituting this time into the exponential produces the required ratio $25/6$ , checking the result.

Answer

$\frac{\ln(25/6)}{0.3}$ hours, equivalently $\frac{10}{3}\ln(25/6)$ hours.

Key idea

The ratio of two exponential readings determines their time separation without requiring the initial amount.
Problem 6: Mark complete
Problem 7 The brightness setting

A setting is defined by $S=\log(L)-2\log(d)$ , where $L>0$ and $d>0$ . The value of $d$ is tripled. By what factor must $L$ change so that $S$ stays the same? Justify the factor.
1. Hint 1
  Combine the logarithms into one logarithm to identify the quantity the setting measures.
2. Hint 2
  The logarithm is unchanged exactly when its positive input is unchanged.
Answer

$L$ must be multiplied by $9$ .

Full solution

The laws apply because $L$ and $d$ are positive.
They give
$S=\log\!\left(\frac{L}{d^2}\right)$

Tripling $d$ multiplies $d^2$ by $9$ , so preserving the positive ratio requires $L$ to be multiplied by $9$ .
Checking,
$\frac{9L}{(3d)^2}=\frac{L}{d^2}$

The argument and therefore the logarithm stay the same.

Answer

$L$ must be multiplied by $9$ .

Key idea

Keeping a logarithmic expression fixed can be reduced to keeping its positive argument fixed.
Problem 7: Mark complete
Problem 8 A sum inside the logarithm

Rae says $\log_2(M)+\log_2(3)$ can never equal $\log_2(M+3)$ for a positive number $M$ . Decide whether Rae is right by finding every positive $M$ that makes the equality true.
1. Hint 1
  Both sides are logarithms to the same base, so ask what has to be true of their inputs.
2. Hint 2
  Combine the left side by the product law and compare the positive logarithm arguments.
Answer

Rae is wrong; $M=\frac32$ .

Full solution

For $M>0$ , the left side is $\log_2(3M)$ and both arguments are positive.
Equal logarithms therefore require
$\begin{aligned}3M&=M+3\\2M&=3\\M&=\frac32\end{aligned}$

This value is positive and makes both arguments $\tfrac92$ , so it works.
The linear equation has no other solution.
The equality is not a general sum law, but it is true at this particular input.

Answer

Rae is wrong; $M=\frac32$ .

Key idea

An invalid general logarithm rule may still give an equality at a particular allowed input.
Problem 8: Mark complete
Problem 9 The negative input

For every real $x<0$ , Arun claims $\log(x^2)=2\log(-x)$ . Is his claim correct? Check that each logarithm is defined for $x<0$ , and justify your decision.
1. Hint 1
  The input to each logarithm must be positive before applying a law.
2. Hint 2
  When $x<0$ , the positive number $-x$ has square $x^2$ .
Answer

Yes, for every $x<0$ .

Full solution

If $x<0$ , then $-x>0$ and $x^2>0$ , so both logarithms exist.
Also
$(-x)^2=x^2$

Apply the power law to the positive input $-x$ :
$\log((-x)^2)=2\log(-x)$

The left side equals $\log(x^2)$ , proving the claim throughout the stated domain.

Answer

Yes, for every $x<0$ .

Key idea

The power law applies to a positive logarithm input, including a positive expression formed from a negative variable.
Problem 9: Mark complete
Problem 10 Jae's comparison

Let $b>0$ with $b\ne1$ , and let $M>0$ . Jae says that if $\log_b(M)=\log_{b^2}(M)$ , then $M=1$ . Is Jae correct? Explain using the meaning of each logarithm.
1. Hint 1
  Give the shared value of the two logarithms a name, and translate each record into an exponential equation.
2. Hint 2
  The bases $b$ and $b^2$ turn the same exponent into powers with exponents differing by a factor of two.
Answer

Yes; $M=1$ .

Full solution

Let the shared value be $q$ .
The definitions give $M=b^q$ and
$M=(b^2)^q=b^{2q}.$
Since $b^x$ is one-to-one,
$q=2q$

Thus $q=0$ , and
$M=b^0$
Since $b^0=1$ , the value is $M=1$ .
Conversely, both logarithms of $1$ are $0$ , so this value does satisfy the equality.

Answer

Yes; $M=1$ .

Key idea

Equal logarithm outputs in related bases can be analyzed by translating both records into powers.
Problem 10: Mark complete

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