where

is the principal amount invested

the amount – principal plus interest - afteryears

is the interest rate for the period in question – 7% would be writteni

is the number of periods.

Something strange happens as the number of periods is increased, and the interest rate per period correspondingly decreased.

If £1000 is invested for 1 years at a fixed rate of 6% per annum the amount after 1 year is

If the amount is paid quarterly four times a year then the interest rate per quarter is 1.5%. The number of periods is now 4. The amount after four years is

Extra interest is earned just by compounding more often even though the interest rate has not gone up.

If the interest is compounded n times a year, then the interest rate per period is 0.06 over n so the amount at the end of the year is

£1000*(1 +{0.06 over n})^n

If we let n tend to infinity- compounding interest every tiny fraction of a nanosecond, we can use the identityto write for the amount after one year,

We can generalise this to any interest rate r and any number of yearsIfis invested for years at a rate of interestcompounded continuously, the amount in the account after n years is

This is continuous compound interest, and always earns more interest than when interest is added after each time period longer than zero.

]]>Example: Solve

By writingwe haveby identifying powers. It is easily seen now that

Example: Solve

Since the base is the same on both sides (it is equal to 3), we can equate the powers to give

If the bases are not the same but are related, we may be able to make them the same.

Example: Solve

We can use one of the indices laws to writethen the equation becomes

The base is the same on both sides so we can equate the bases, obtainingThis can easily be solved:

If the bases are not related, we can still solve the equation, but we must take logs.

Example: Solve

Taking logs gives

Expand the brackets and collect coefficients of

Now divide bythe coefficient ofto give

This can be evaluated by calculator, but this does not give an exact answer. We can instead make a single log of numerator and denominator, obtaining.

Example: Solve

This has no solutions. We can raise a real number to a negative power, but the result, if the power is real, can never be a negative number.

Example: Solve

Substitutethenand we obtain

We can factorise this to obtainthenor

Use the substitution now to obtainor

]]>To find the intersections with the axis, put each coordinate equal to 0.

Ifwe find the intersection with the– axis by putting

We find the– intersection by puttingThis has no real solution but since tends to 0 fortending towe can take the– intercept to be at

To find the equation of the asymptote(s) putandsuccessively equal to

Puttingimpliesand vice versa, so this gives us no asymptote. Put to getPuttingreturns nothing forsincehas no solution for

Ifwe find the intersection with the– axis by putting

We find the– intersection by putting

To find the equation of the asymptote(s) putandsuccessively equal to

Puttingimpliesand vice versa, so this gives us no asymptote. Put to getPuttingreturns nothing forsincehas no solution for

]]>To multiply numbers with the same base, keep the base and add the indices.

To divide numbers with the same base, keep the base and subtract the indices.

When raising a power to a power, keep the base and multiply the indices.

The power of a product is the product of the powers.

A power of a quotient is the quotient of the powers.

The root of a quotient is the quotient of the roots .

The root of a quotient raised to a power is the quotient of the roots of the powers.

(is not defined)

Any non-zero number raised to the power of zero is 1.

and

A negative exponent means take the reciprocal.

A root is a fractional power. Taking the nth root is the same as raising to the power

Using the third and seventh rules allows us to raise a root to a power.

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We can write the first of the as {jatex options:inline}10^2 \times 10^x=10^{2y} \rightarrow 2+x=2y{/jatex}

We can write the second of these as {jatex options:inline}\frac{10^1 \times 10^x}{10^y}=10^2 \rightarrow 1+x-y=2{/jatex}

We now have the equations

{jatex options:inline}2+x=2y{/jatex} (1)

{jatex options:inline}1+x-y=2{/jatex} (2)

Thje first of these minus the second gives {jatex options:inline}(2+x)-(1+x-y)=2y-2 \rightarrow 1+y=2y-2 \rightarrow 1+2=2y-y \rightarrow y=3{/jatex}.

Substiture {jatex options:inline}y=3{/jatex} into (1) then {jatex options:inline}x=2y-2=4{/jatex}]]>

The above graphs ofandare the graphcompressed in the– direction. For example we can writeThe graph is compressed in the- direction by the factor

The above graphs are the reflections of the graphs graphsandin the- axis. The same transformation is achieved by making the swap

The transformation above are translations of the exponential curveby 2 to the left to giveand a translation of 1 to the right to giveNotice that a negative- transformation by two becomes ain the equation of the graph, and a positive translation of bybecomes ain the equation of the graph. This is because these particular transformations are– transformations, and– transformations are always counter intuitive.

The graph ofabove is the reflection of the graphin the– axis. This simple introduces a factor ofin the equation ofto give

The graphis transformed above by moving up by 3 and down by 1 (– transformations) respectively. The equation becomesandrespectively.

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