Back to homeMathematics · High School Year 2 · Physics-Math · Change subject
Start for free
Prep 2 Tech · Mathematics
Real numbers · Real numbers

Scientific Notation

Before you start, make sure you know: Properties of powers (High School Year 1 (semesters 1 and 2))
Before you start, make sure you know: Powers with integer exponents (High School Year 1 (semesters 1 and 2))

Your cell phone performs billions of operations per second, and the mass of an electron is $9.11 \times 10^{-31}$ kg. Written in decimal form, it would be a decimal point followed by $30$ zeros and then $911$. Who wants to write that many zeros? Scientific notation is the language that physics, chemistry, and engineering use to handle huge or tiny numbers without getting lost.

Mantissa and order of magnitude

Practice★☆☆Not started
Write the following numbers in standard form, adjusting the mantissa so that it lies between $1$ and $10$: a) $0.000000456$ b) $34500000000$ c) $0.0023 \times 10^5$ d)…

Write the following numbers in standard form, adjusting the mantissa so that it lies between $1$ and $10$:

  1. a) $0.000000456$
  2. b) $34500000000$
  3. c) $0.0023 \times 10^5$
  4. d) $456 \times 10^{-4}$
Solve
Practice★☆☆Not started
Justify whether the following expressions are correctly written in standard form. If not, correct them so that the mantissa $N$ satisfies $1 \le |N| < 10$: a)…

Justify whether the following expressions are correctly written in standard form. If not, correct them so that the mantissa $N$ satisfies $1 \le |N| < 10$:

  1. a) $12.4 \times 10^3$
  2. b) $0.85 \times 10^{-2}$
  3. c) $1 \times 10^6$
  4. d) $-45.2 \times 10^{-8}$
Solve

Converting between decimal and scientific form

Two conversions: 0.0075 mm to 7.5 times 10 to the minus 6 meters and 185 billion to 1.85 times 10 to the 11 0.0075 mm = 7.5 · 10⁻⁶ m 3 places for going from mm to m, plus 3 for the point: 6 to the left 185 billion = 1.85 · 10¹¹ 185 000 · 10⁶ = 1.85 · 10⁵ · 10⁶
Practice★☆☆Not started
The diameter of a human red blood cell is approximately $0.0075$ mm. Express this quantity in meters using scientific notation.

The diameter of a human red blood cell is approximately $0.0075$ mm. Express this quantity in meters using scientific notation.

Solve
Practice★☆☆Not started
The estimated wealth of one of the richest people in the tech sector amounts to $185000$ million dollars. Write this figure in standard form.

The estimated wealth of one of the richest people in the tech sector amounts to $185000$ million dollars. Write this figure in standard form.

Solve

Significant figures and rounding

The significant figures are the digits that carry information in a measurement. In scientific notation they are easy to count: they are all the digits of the mantissa, including trailing zeros (because someone took the trouble to measure them).

Practice★☆☆Not started
State the order of magnitude and the number of significant figures of the following experimental measurements: a) $4.05 \times 10^8$ m b) $1.200 \times 10^{-5}$ kg c)…

State the order of magnitude and the number of significant figures of the following experimental measurements:

  1. a) $4.05 \times 10^8$ m
  2. b) $1.200 \times 10^{-5}$ kg
  3. c) $9 \times 10^{12}$ s
Solve
Practice★☆☆Not started
Round the following numbers to three significant figures and then express them in standard form: a) $4567892$ b) $0.00034091$ c) $98.99$

Round the following numbers to three significant figures and then express them in standard form:

  1. a) $4567892$
  2. b) $0.00034091$
  3. c) $98.99$
Solve

Comparing numbers in scientific notation

Number line with three consecutive marks of powers of 10 and two numbers placed between them 10⁴10⁵10⁶ 1.5·10⁴3.2·10⁵
Practice★★☆Not started
Order the following numbers from smallest to largest, first writing them in standard form and comparing their orders of magnitude: $A = 0.0045 \times 10^{-3}$…

Order the following numbers from smallest to largest, first writing them in standard form and comparing their orders of magnitude:

$A = 0.0045 \times 10^{-3}$

$B = 450 \times 10^{-8}$

$C = 0.45 \times 10^{-5}$

$D = 45000 \times 10^{-11}$

Solve

Multiplying and dividing

Multiplication in scientific notation: the mantissas are multiplied and the exponents are added 6 · 10⁸ · 4 · 10⁻³ = 24 · 10⁵ = 2.4 · 10⁶ mantissas: multiplied exponents: added the mantissa is readjusted

To multiply, you operate on the mantissas and the powers of $10$ separately (the exponents are added). To divide, you divide the mantissas and subtract the exponents. At the end, if the mantissa falls outside $[1, 10)$, you readjust it.

Practice★☆☆Not started
Carry out the following multiplications and divisions mentally (without using a calculator), applying the index laws, and express the result in standard form: a)…

Carry out the following multiplications and divisions mentally (without using a calculator), applying the index laws, and express the result in standard form:

  1. a) $(3 \times 10^4) \times (2 \times 10^5)$
  2. b) $\frac{8 \times 10^{-3}}{4 \times 10^{-7}}$
  3. c) $(1.5 \times 10^6) \times (4 \times 10^{-2})$
Solve
Practice★★☆Not started
The mass of the Earth is approximately $5.97 \times 10^{24}$ kg and that of the planet Jupiter is $1.90 \times 10^{27}$ kg. How many times more massive is Jupiter than…

The mass of the Earth is approximately $5.97 \times 10^{24}$ kg and that of the planet Jupiter is $1.90 \times 10^{27}$ kg. How many times more massive is Jupiter than the Earth? Express the result in standard form and also in standard decimal form.

Solve
Practice★★☆Not started
The speed of light in a vacuum is approximately $3 \times 10^5$ km/s. A "light-year" is the distance light travels in a vacuum in one Earth year ($365.25$ days).…

The speed of light in a vacuum is approximately $3 \times 10^5$ km/s. A "light-year" is the distance light travels in a vacuum in one Earth year ($365.25$ days). Calculate how many kilometers a light-year corresponds to, and express the result in scientific notation to four significant figures.

Solve

Adding and subtracting

Addition with different exponents: one of the numbers is rewritten to match exponents and the common factor is factored out 6.1·10⁶ + 2.5·10⁷ = 0.61·10⁷ + 2.5·10⁷ = (0.61 + 2.5)·10⁷ = 3.11·10⁷
Practice★☆☆Not started
Perform the following additions and subtractions of numbers in standard form that have the same exponent. Factorise out a common factor and give the final result: a)…

Perform the following additions and subtractions of numbers in standard form that have the same exponent. Factorise out a common factor and give the final result:

  1. a) $3.2 \times 10^5 + 4.5 \times 10^5$
  2. b) $8.7 \times 10^{-4} - 3.2 \times 10^{-4}$
  3. c) $5.15 \times 10^{12} + 2.03 \times 10^{12}$
Solve
Practice★★☆Not started
Perform the following additions and subtractions, first adjusting the exponents so you can factorise out a common factor. Express the final result in standard form: a)…

Perform the following additions and subtractions, first adjusting the exponents so you can factorise out a common factor. Express the final result in standard form:

  1. a) $4.5 \times 10^7 + 3.8 \times 10^8$
  2. b) $9.2 \times 10^{-5} - 6.1 \times 10^{-6}$
Solve
Practice★★☆Not started
Calculate the following mixed operation and express the final result in standard form, rounding to three significant figures:…

Calculate the following mixed operation and express the final result in standard form, rounding to three significant figures:

$\frac{(2.5 \times 10^6) \times (4.2 \times 10^{-3})}{3.1 \times 10^{-4} + 1.9 \times 10^{-5}}$

Solve

Powers, roots, and percent growth

The properties of powers apply: $(a \times 10^n)^k = a^k \times 10^{nk}$. For the square root, it helps to have an even exponent, because $\sqrt{10^{2m}} = 10^m$.

If a quantity grows $r\,\%$ each year, it is multiplied by the factor $1 + \frac{r}{100}$ each year. After $t$ years, it is multiplied by $\left(1 + \frac{r}{100}\right)^t$.

Practice★★☆Not started
A server hard drive has a storage capacity of $2$ TB (terabytes). Given that $1\text{ TB} = 1024\text{ GB}$, $1\text{ GB} = 1024\text{ MB}$,…

A server hard drive has a storage capacity of $2$ TB (terabytes). Given that $1\text{ TB} = 1024\text{ GB}$, $1\text{ GB} = 1024\text{ MB}$, $1\text{ MB} = 1024\text{ KB}$ and $1\text{ KB} = 1024\text{ B}$ (bytes), calculate the exact number of bytes the hard drive can store and express the result in standard form.

Solve
Practice★★★Not started
Given the physical quantities $A = 3.2 \times 10^5$ and $B = 5.1 \times 10^{-4}$, calculate the value of the algebraic expression $\frac{A^3}{\sqrt{B}}$. Express the…

Given the physical quantities $A = 3.2 \times 10^5$ and $B = 5.1 \times 10^{-4}$, calculate the value of the algebraic expression $\frac{A^3}{\sqrt{B}}$. Express the final result in standard form to three significant figures and clearly state its order of magnitude.

Solve
Practice★★★Not started
Global Gross Domestic Product (GDP) in a given year was approximately $170 \times 10^{13}$ pesos. If the global population at that time was $7.8 \times 10^9$…

Global Gross Domestic Product (GDP) in a given year was approximately $170 \times 10^{13}$ pesos. If the global population at that time was $7.8 \times 10^9$ inhabitants, calculate the mean GDP per capita. If the population is projected to grow by $1.2\%$ per year and GDP is projected to grow by $2.5\%$ per year, estimate the GDP per capita after $10$ years. Express all results in scientific notation.

Solve

Common logarithms and exponentials

A common logarithm is split into a whole part, which is the exponent, and a decimal part, which gives the mantissa log N = 77 . 06368 whole part → exponent of 10 decimal part → mantissa = 10^0.06368 ≈ 1.158 N ≈ 1.158 · 10⁷⁷

When a number is so large that the calculator won't display it, like $2^{256}$, logarithms help. If $\log_{10} N = k + f$, with $k$ an integer and $0 \le f < 1$, then $N = 10^{f} \times 10^{k}$, and $10^f$ lies between $1$ and $10$: it is already the mantissa.

When the formula has an exponential, like $N_0 e^{-\lambda t}$, you compute the exponent with a calculator and then convert to scientific notation.

Practice★★★Not started
A high-security encryption algorithm uses $256$-bit keys, which means the total number of possible keys is $2^{256}$. Using the properties of common logarithms (base 10)…

A high-security encryption algorithm uses $256$-bit keys, which means the total number of possible keys is $2^{256}$. Using the properties of common logarithms (base 10) and given that $\log_{10} 2 \approx 0.30103$, express the total number of combinations in standard form $N \times 10^n$, where $1 \le N < 10$.

Solve
Practice★★★Not started
A sample of radioactive material used in fossil dating initially contains $N_0 = 4.5 \times 10^{20}$ nuclei. The number of nuclei decays exponentially according to the…

A sample of radioactive material used in fossil dating initially contains $N_0 = 4.5 \times 10^{20}$ nuclei. The number of nuclei decays exponentially according to the law $N(t) = N_0 \times e^{-\lambda t}$, where $\lambda = 3.2 \times 10^{-4}\text{ years}^{-1}$. Calculate the number of undecayed nuclei remaining after $5000$ years. Express the final result in standard form to two significant figures.

Solve
Practice★★★Not started
The energy released by an earthquake is empirically related to its magnitude $M$ on the Richter scale by the formula $\log_{10} E = 11.8 + 1.5M$, where $E$ is measured…

The energy released by an earthquake is empirically related to its magnitude $M$ on the Richter scale by the formula $\log_{10} E = 11.8 + 1.5M$, where $E$ is measured in ergs. Calculate the energy released by an earthquake of magnitude $7.5$ and by another of magnitude $8.5$. How many times greater is the energy of the second compared to the first? Express the energy results in standard form.

Solve

Measurement errors and intervals

Number line with the true value and the approximate value, and the distance between them as the absolute error true value approximation absolute error

If a measurement lies in an interval $[a, b]$, the natural estimated value is the midpoint $\frac{a+b}{2}$. The bound on the absolute error is half the length of the interval, $\frac{b-a}{2}$.

Practice★★☆Not started
When measuring the distance between two cities using a GPS system, a real value of $456321$ m is obtained. A tourist map approximates this distance, giving a value of…

When measuring the distance between two cities using a GPS system, a real value of $456321$ m is obtained. A tourist map approximates this distance, giving a value of $4.56 \times 10^5$ m. Calculate the absolute error and the relative error made when using the map's approximation. Express both errors in standard form.

Solve
Practice★★★Not started
The edge of a perfect cube has been measured with a precision instrument and its length is known to lie in the interval $[1.25 \times 10^{-2}, 1.26 \times 10^{-2}]$ m.…

The edge of a perfect cube has been measured with a precision instrument and its length is known to lie in the interval $[1.25 \times 10^{-2}, 1.26 \times 10^{-2}]$ m. Determine the interval in which the volume of the cube lies, expressed in $\text{mm}^3$, giving the end points in standard form. Calculate the absolute error bound if we take the midpoint of this interval as the estimated volume.

Solve

Applications in science and engineering

Calculation of the force of gravity between the Earth and the Moon with data in scientific notation F = G · m₁ · m₂ / d² G = 6.67·10⁻¹¹ · m₁ = 5.97·10²⁴ · m₂ = 7.35·10²² · d = 3.84·10⁸ m F ≈ 1.98 · 10²⁰ N
Practice★★☆Not started
Given that $18$ grams of water contain $6.022 \times 10^{23}$ molecules (Avogadro's number), calculate the mass in grams of a single molecule of water. Give the result…

Given that $18$ grams of water contain $6.022 \times 10^{23}$ molecules (Avogadro's number), calculate the mass in grams of a single molecule of water. Give the result in standard form to four significant figures.

Solve
Practice★★★Not started
The total volume of water in all Earth's oceans is estimated at $1.332 \times 10^9\text{ km}^3$. Assuming that a standard drop of water has an average volume of $0.05$…

The total volume of water in all Earth's oceans is estimated at $1.332 \times 10^9\text{ km}^3$. Assuming that a standard drop of water has an average volume of $0.05$ ml, determine the total number of drops of water in the oceans. Express the result in standard form. (Remember that $1\text{ ml} = 1\text{ cm}^3$).

Solve
Practice★★★Not started
Newton's law of universal gravitation states that the gravitational force of attraction between two masses is $F = G \frac{m_1 m_2}{d^2}$. Given that the constant…

Newton's law of universal gravitation states that the gravitational force of attraction between two masses is $F = G \frac{m_1 m_2}{d^2}$. Given that the constant $G = 6.67 \times 10^{-11}\text{ N}\times\text{m}^2/\text{kg}^2$, the mass of the Earth is $m_1 = 5.97 \times 10^{24}$ kg, the mass of the Moon is $m_2 = 7.35 \times 10^{22}$ kg and the mean distance between their centers is $d = 3.84 \times 10^5$ km, calculate the force of attraction $F$ in newtons, using standard form in all intermediate calculations.

Solve

Back to the beginning: the mass of the electron is $9.11 \times 10^{-31}$ kg, and the distance to Proxima Centauri is about $4.02 \times 10^{16}$ m. With scientific notation both fit on one line and can be multiplied or compared without counting zeros.

Summary

See all the exercises in this section