1.9 - Standard form

1.9 - Standard form

Standard form makes very large and very small numbers manageable. Convert and calculate with powers of ten, read calculator displays and, at Higher tier, build a model for a problem with large quantities.

Interpreting standard form

An archive has 1.6×1081.6\times10^8 records, each using 2.5×1032.5\times10^3 bytes. Could a link sending 8.0×1068.0\times10^6 bytes each second transfer it in one hour? Those powers look fierce, but they keep the zeros under control. We will return to this question.

You already know

103=100010^3=1000, 100=110^0=1 and 10−3=1/100010^{-3}=1/1000. When multiplying powers of ten, add exponents; when dividing, subtract them.

Start with an ordinary large number: 5 820 0005\,820\,000 is 5.825.82 million. Standard form writes “million” as 10610^6.

Standard form

A positive number is in standard form when it is a×10na\times10^n, where 1≤a<101\leq a<10 and nn is an integer. The coefficient aa supplies the significant digits; the exponent nn sets the scale.

For example, 5.82×1065.82\times10^6 is in standard form. The equal number 58.2×10558.2\times10^5 is not, because its coefficient is too large. Zero does not fit this positive-number definition. The definition and conversion rules are not in either tier's formulae sheet: learn them.

Worked example 1: a large ordinary number

Question. Write 5 820 0005\,820\,000 in standard form.

Plan. Choose a coefficient between 11 and 1010, then restore the original scale.

5 820 000=5.82×1 000 000separate digits and scale=5.82×106a million is 106\begin{aligned} 5\,820\,000&=5.82\times1\,000\,000 &&\text{separate digits and scale}\\ &=5.82\times10^6 &&\text{a million is }10^6 \end{aligned}

Why a positive exponent? Multiplying the small coefficient by a large power restores the large original number.

Answer. 5.82×1065.82\times10^6.

Check. 1≤5.82<101\leq5.82<10, and multiplying by a million returns 5 820 0005\,820\,000.

Now change the scale to a number below one. The exponent must divide the coefficient by a power of ten.

Worked example 2: a small ordinary number

Question. Write 0.0000730.000073 in standard form.

Plan. Use coefficient 7.37.3, then find the divisor that returns the original number.

0.000073=7.3100 000restore the small size=7.3×10−5divide by 105\begin{aligned} 0.000073&=\frac{7.3}{100\,000} &&\text{restore the small size}\\ &=7.3\times10^{-5} &&\text{divide by }10^5 \end{aligned}

Why −5-5, rather than 55? A positive power would give 730 000730\,000, which is much too large.

Answer. 7.3×10−57.3\times10^{-5}.

Check. Dividing 7.37.3 by 100 000100\,000 gives 0.0000730.000073.

Reverse the process to recover ordinary numbers: 8.4×10−3=8.4/1000=0.00848.4\times10^{-3}=8.4/1000=0.0084, while 4.7×100=4.74.7\times10^0=4.7. A zero exponent does not make the number zero.

Watch out

The line 0.000073=7.3×1050.000073=7.3\times10^5 reverses the scale. Use 7.3×10−57.3\times10^{-5}; always compare your answer's size with the original.

Try a small number and a reverse conversion together.

Comparing, adding and subtracting

Addition and subtraction need the same scale, just as money amounts need the same currency.

For positive numbers already in standard form, compare exponents first. If they match, compare coefficients. Thus 4.9×10−4<1.2×10−34.9\times10^{-4}<1.2\times10^{-3}: 0.000490.00049 is less than 0.00120.0012, despite having the larger coefficient. A larger integer exponent gives the larger scale, including when both exponents are negative.

To combine amounts, rewrite one using the other's power of ten. The temporary coefficient may fall outside the standard-form range; that is fine during working. This place-value method is not in either tier's formulae sheet.

Worked example 3: adding at a common scale

Question. Work out 3.6×105+8.4×1043.6\times10^5+8.4\times10^4 in standard form.

Plan. Rewrite both terms as multiples of 10510^5.

8.4×104=0.84×105same value, common scale3.6×105+8.4×104=(3.6+0.84)×105collect equal units=4.44×105add coefficients\begin{aligned} 8.4\times10^4&=0.84\times10^5 &&\text{same value, common scale}\\ 3.6\times10^5+8.4\times10^4&=(3.6+0.84)\times10^5 &&\text{collect equal units}\\ &=4.44\times10^5 &&\text{add coefficients} \end{aligned}

Why not add 3.63.6 and 8.48.4? They originally count different sizes of unit.

Answer. 4.44×1054.44\times10^5.

Check. 360 000+84 000=444 000360\,000+84\,000=444\,000.

Change only the operation. Keep the common scale, but subtract the aligned coefficients.

Worked example 4: subtracting at the same scale

Question. Work out 3.6×105−8.4×1043.6\times10^5-8.4\times10^4 in standard form.

Plan. Use the same rewrite, then subtract.

8.4×104=0.84×105align powers3.6×105−8.4×104=(3.6−0.84)×105collect equal units=2.76×105subtract coefficients\begin{aligned} 8.4\times10^4&=0.84\times10^5 &&\text{align powers}\\ 3.6\times10^5-8.4\times10^4&=(3.6-0.84)\times10^5 &&\text{collect equal units}\\ &=2.76\times10^5 &&\text{subtract coefficients} \end{aligned}

Why keep the exponent? Subtraction changes how many units there are, not the size of the common unit.

Answer. 2.76×1052.76\times10^5.

Check. 360 000−84 000=276 000360\,000-84\,000=276\,000, smaller than the starting number.

The method is unchanged for negative exponents. For example, 7.2×10−3−4.8×10−4=(7.2−0.48)×10−3=6.72×10−37.2\times10^{-3}-4.8\times10^{-4}=(7.2-0.48)\times10^{-3}=6.72\times10^{-3}.

Watch out

The line 2.5×104+3.2×103=5.7×1072.5\times10^4+3.2\times10^3=5.7\times10^7 is wrong. Adding exponents belongs to multiplication, not addition. Align scales instead: (2.5+0.32)×104=2.82×104(2.5+0.32)\times10^4=2.82\times10^4.

Keep the subtraction, but now work below one.

Multiplying and dividing

Products and quotients let you handle significant digits and scale separately.

Standard-form arithmetic

For integer exponents m,nm,n,

(a×10m)(b×10n)=ab×10m+n(a\times10^m)(b\times10^n)=ab\times10^{m+n} a×10mb×10n=ab×10m−n(b≠0)\frac{a\times10^m}{b\times10^n}=\frac ab\times10^{m-n}\qquad(b\ne0)

Not in the formulae sheet: learn it. Neither the Foundation nor Higher sheet supplies these index laws.

These follow from regrouping factors and the index laws. If the resulting coefficient is outside 1≤a<101\leq a<10, normalise it: transfer a factor of ten between coefficient and power without changing the value.

Worked example 5: multiplying and normalising

Question. Calculate (6×107)(4×10−3)(6\times10^7)(4\times10^{-3}) in standard form.

Plan. Multiply coefficients, add exponents, then inspect the coefficient.

(6×107)(4×10−3)=(6×4)×107+(−3)regroup factors=24×104coefficient and power=2.4×105normalise\begin{aligned} (6\times10^7)(4\times10^{-3})&=(6\times4)\times10^{7+(-3)} &&\text{regroup factors}\\ &=24\times10^4 &&\text{coefficient and power}\\ &=2.4\times10^5 &&\text{normalise} \end{aligned}

Why increase the exponent? Dividing the coefficient by 1010 needs a compensating multiplication by 1010.

Answer. 2.4×1052.4\times10^5.

Check. 60 000 000×0.004=240 00060\,000\,000\times0.004=240\,000.

Change only multiplication to division. A negative denominator exponent must be subtracted with its sign intact.

Worked example 6: dividing by a small number

Question. Calculate (6×107)÷(4×10−3)(6\times10^7)\div(4\times10^{-3}) in standard form.

Plan. Divide coefficients and subtract the whole denominator exponent.

6×1074×10−3=64×107−(−3)quotient law=1.5×1010subtract a negative\begin{aligned} \frac{6\times10^7}{4\times10^{-3}}&=\frac64\times10^{7-(-3)} &&\text{quotient law}\\ &=1.5\times10^{10} &&\text{subtract a negative} \end{aligned}

Why does the exponent grow? Dividing by 0.0040.004 makes the original quantity much larger.

Answer. 1.5×10101.5\times10^{10}.

Check. (1.5×1010)(4×10−3)=6×107(1.5\times10^{10})(4\times10^{-3})=6\times10^7, recovering the dividend.

A coefficient can also be too small. For example, (1.8×10−4)/(3×102)=0.6×10−6=6×10−7(1.8\times10^{-4})/(3\times10^2)=0.6\times10^{-6}=6\times10^{-7}. Multiplying the coefficient by 1010 requires reducing the exponent by 11.

Watch out

The line 107−(−3)=10410^{7-(-3)}=10^4 loses the denominator's negative sign. Write 7−(−3)=107-(-3)=10, then use 101010^{10}.

If Pearson asks for standard form, the final coefficient matters as well as the numerical value.

Exam tip

24×10424\times10^4 has the correct value but is not in standard form. Finish the normalisation when that form is requested.

Try another quotient whose first coefficient is below one.

Calculator notation and precision

A calculator's scientific display uses the same powers of ten, but its screen notation is not your final written answer.

A display such as 5.82E65.82\mathrm{E}6 means 5.82×1065.82\times10^6. The E is a scale marker. The key labelled EXP, EE or a power of ten supplies that factor already: enter the coefficient, that key, then the exponent. Do not add another multiplication by 1010. Key names vary, so check your own device's display against a known value.

Retain full stored precision during calculations. Round once at the end to the stated accuracy. In a standard-form number, round the coefficient's significant figures, then check the allowed coefficient range again. This rounding procedure is not on either tier's formulae sheet.

Worked example 7: rounding the coefficient

Question. Give 9.86×1089.86\times10^8 to two significant figures in standard form.

Plan. Round the coefficient, then check its size.

9.86≈9.9two significant figures9.86×108≈9.9×108keep the scale\begin{aligned} 9.86&\approx9.9 &&\text{two significant figures}\\ 9.86\times10^8&\approx9.9\times10^8 &&\text{keep the scale} \end{aligned}

Why keep the exponent? The rounded coefficient still lies between 11 and 1010.

Answer. 9.9×1089.9\times10^8.

Check. The rounding changes 986 000 000986\,000\,000 to 990 000 000990\,000\,000, a small increase.

Change only the coefficient so that rounding crosses the upper boundary. The value must be normalised again.

Worked example 8: rounding across a power of ten

Question. Give 9.96×1089.96\times10^8 to two significant figures in standard form.

Plan. Round first, then replace the coefficient 1010 with a valid coefficient.

9.96≈10two significant figures10×108=1.0×109normalise and show precision\begin{aligned} 9.96&\approx10 &&\text{two significant figures}\\ 10\times10^8&=1.0\times10^9 &&\text{normalise and show precision} \end{aligned}

Why write 1.01.0? The final zero shows that two significant figures are being reported.

Answer. 1.0×1091.0\times10^9.

Check. 996 000 000996\,000\,000 rounds to 1 000 000 0001\,000\,000\,000 to two significant figures.

A small coefficient change can change the written exponent, even though the actual numbers remain close. That is a feature of the notation, not a sudden jump in size.

Watch out

The answer 10×10810\times10^8 has the rounded value but fails a<10a<10. Write 1.0×1091.0\times10^9, preserving both value and requested precision.

Make the same boundary crossing at a much smaller scale.

Choosing a standard-form model (Higher)

In a word problem, the units help you choose the calculation before you touch the exponents.

A total amount is the number of items multiplied by the amount per item. A time at a constant rate is amount divided by amount per unit time. These relationships are not supplied on either tier's formulae sheet. Read the stated assumptions: a constant transfer rate and no overheads are a model, not a promise about every real network.

Worked example 9: an archive transfer

Question. An archive contains 1.6×1081.6\times10^8 records of 2.5×1032.5\times10^3 bytes each. A link transfers 8.0×106 bytes/s8.0\times10^6\,\mathrm{bytes/s} at a constant rate, with no overheads. Find the transfer time in minutes, in standard form to three significant figures. Use 6060 seconds per minute.

Plan. Find total bytes, divide by bytes per second, then convert seconds to minutes.

D=(1.6×108)(2.5×103)records times bytes each=4.0×1011 bytestotal datat=4.0×10118.0×106divide by rate=5.0×104 sbytes cancelt=50 00060=833.333… minconvert time\begin{aligned} D&=(1.6\times10^8)(2.5\times10^3) &&\text{records times bytes each}\\ &=4.0\times10^{11}\,\mathrm{bytes} &&\text{total data}\\ t&=\frac{4.0\times10^{11}}{8.0\times10^6} &&\text{divide by rate}\\ &=5.0\times10^4\,\mathrm{s} &&\text{bytes cancel}\\ t&=\frac{50\,000}{60}=833.333\ldots\,\mathrm{min} &&\text{convert time} \end{aligned}

Why divide by the rate? Each second removes that many bytes, so total bytes divided by bytes per second counts seconds.

Answer. 8.33×102 min8.33\times10^2\,\mathrm{min}.

Check. 50 000×8.0×106=4.0×101150\,000\times8.0\times10^6=4.0\times10^{11} bytes. The exact time is far more than one hour.

Change only the rate: a faster link should reduce the time. Predict the direction before recalculating.

Worked example 10: a faster link

Question. Transfer the same archive at 1.6×107 bytes/s1.6\times10^7\,\mathrm{bytes/s}, with the same assumptions. Find the time in minutes, in standard form to three significant figures.

Plan. Keep the total data and divide by the new rate.

t=4.0×10111.6×107new rate only=2.5×104 sdivide coefficients and powerst=25 00060=416.666… minconvert last\begin{aligned} t&=\frac{4.0\times10^{11}}{1.6\times10^7} &&\text{new rate only}\\ &=2.5\times10^4\,\mathrm{s} &&\text{divide coefficients and powers}\\ t&=\frac{25\,000}{60}=416.666\ldots\,\mathrm{min} &&\text{convert last} \end{aligned}

Why keep all the seconds? Rounding before the unit conversion could alter the final significant figures.

Answer. 4.17×102 min4.17\times10^2\,\mathrm{min}.

Check. The rate doubled, and the unrounded time halved. It still exceeds one hour.

There is a useful verified surprise in this model: doubling the rate halves the time exactly, regardless of how many zeros the data count has. If instead you ask how many whole files fit on a drive, divide capacity by file size and round down only at the final whole-file decision.

Watch out

The line t=Drt=Dr is wrong when rr is bytes per second. Its units are not time. Use t=D/rt=D/r, which gives seconds.

The opening archive therefore cannot transfer in one hour under the stated model: it takes 50 000 s50\,000\,\mathrm{s}, while an hour has 3600 s3600\,\mathrm{s}.

Quick Recap

  • Standard form is a×10na\times10^n, with 1≤a<101\leq a<10 and integer nn.
  • Align powers before adding or subtracting; compare normalised exponents before coefficients.
  • Multiply coefficients and add exponents; divide coefficients and subtract exponents.
  • Normalise after arithmetic or rounding, and write mathematical notation instead of a calculator E display.
  • Use units to choose a model and keep full precision until the final answer.

The opening archive takes 50 000 s50\,000\,\mathrm{s}, or about 833833 minutes, so the one-hour transfer is impossible under the stated assumptions.