Question 4.13: Simplify each of the following expressions, giving your answ......

(a) {\frac{2^{5}}{2^{3}2^{-4}}} = {\frac{2^{5}}{2^{3  –  4}}}        use rule 1 to add the indices of the base 2 terms below the line

=  {\frac{2^{5}}{2^{-1}}}        simplify

= 2^{5−(−1)}        use rule 2 and bracket the negative terms

= 2^{5  + 1} = 2^6      use the calculator to evaluate exponentials

= 64

(b) Q Q^{1.5}P = Q¹ Q^{1.5}P        since no index is given on the first Q, it is understood to be 1

= Q^{1+1.5}P

= Q^{2.5}P

(c) ~\frac{{\sqrt{3^{5}}}}{3^{-4}3^{4}} = \frac{3^{5/2}}{3^{4-4}}      start by writing \sqrt{3^{5}} = (3^5)^{1/2} = (3^{5/1})^{1/2}

= \frac{3^{5/2}}{3^0}        = 3^{5/1×1/2} = 3^{5/2}

= 3^{5/2}        since 3^0 = 1

=15.588

(d) {\frac{15e^{\alpha}e^{\beta}}{e^{2}}} = {\frac{15e^{\alpha + \beta}}{e^{2}}}      multiplying numbers with same base (rule 1)

= \frac{15e^{α + β} e^{-2}}{1}    bringing up the number from below the line (note 2)

= 15 e^{α + β  –  2}    multiplying numbers with same base (rule 1)

(e) \left({\frac{3L^{0.5}}{L^{-2}}}\right)^{2} = (\frac{3L^{0.5  –  (-2)}}{1})^2      simplify the terms inside the bracket first

= (\frac{3L^{0.5 + 2}}{1})^2

= (\frac{3L^{2.5}}{1})^2

= (\frac{3L^{2.5}}{1}) (\frac{3L^{2.5}}{1})        square out

= (\frac{3^2L^{2.5 + 2.5}}{1})

= 9L^5

There are usually several different approaches to simplifying; for example, in part (e) we could square out the brackets first then simplify like this:

\left({\frac{3^{1}L^{0.5}}{L^{-2}}}\right)^{2} = {\frac{3^{1\times2}L^{0.5\times2}}{L^{-2\times2}}}      square first by multiplying the index on every term   (including the 3) by the power 2 outside the bracket

= ={\frac{9L}{L^{-4}}}

= {\frac{9L^{1-(-4)}}{1}}

= 9L^{5}

(f) \left({\frac{3e^{0.5}}{e^{-2}}}\right)^{2} = \left({\frac{3e^{0.5}}{e^{-2}}}\right) \left({\frac{3e^{0.5}}{e^{-2}}}\right)      squaring

= \frac{3^{2}(e^{0.5})^{2}}{(e^{-2})^{2}}

= {\frac{3^{2}e^{1}}{e^{-4}}}      multiplying the indices (rule 3)

= 9e^{1-(-4)}        subtracting the index of the divisor (rule 2)

= 9e^{5}=1335.7184      simplifying, then evaluating correct to 4 decimal places

(g) The rules for indices do not apply to sums or differences of the exponential functions. But this expression may be simplified by factorisation, as follows:

\frac{p^5q  −  p^4q²}{p^5  –  p^4q } = \frac{p^{4}p q  –  p^{4}q q}{p^{4}p  –  p^{4}q}    common factors p^4q in the numerator and p^4 in the denominator

= \frac{p^{4}(p-q)q}{p^{4}(p-q)}    factor out the common factors

= \frac{\cancel{p^{4}}\cancel{(p-q)}q}{\cancel{p^{4}}\cancel{(p-q)}} = q    cancel the factors that are identical above and below the line

(h) \frac{4e^{\alpha+2\beta}-2e^{\alpha}}{2e^{\alpha}} = \frac{4e^{\alpha}e^{2\beta}-2e^{\alpha}}{2e^{\alpha}}      rule 1: e^{α+2β} = e^αe^{2β}

= \frac{2e^{\alpha}(2e^{2\beta}-1)}{2e^{\alpha}}    common factor 2e^α in the numerator

= 2e^{2\beta}-1      cancel the factor that is common in the numerator and denominator

Alternatively

\frac{4e^{\alpha+2\beta}-2e^{\alpha}}{2e^{\alpha}}= \frac{4e^{\alpha+2\beta}}{2e^{\alpha}}  –  \frac{2e^{\alpha}}{2e^{\alpha}} = \frac{4e^{\alpha+2\beta-\alpha}}{2}  –  \frac{2e^{\alpha}}{2e^{\alpha}} = 2e^{2\beta}-1