Maths Worksheet 3 - outline answers PDF

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Maths for Chemists

2016

Workshop 3 Logarithms, Exponentials, Trigonometric identities. Outline answers

Note: there are often several methods for solving the same mathematical problem. Choose whichever you like, but make sure to show your working.

1.

2.

(i)

 (3 2 )(2 4 )  2 ln 3 4 ln 2 3ln 5 ln(32 )  ln(24 )  ln(53 )  ln   3  (5 )  144   9  16   ln   ln    125  125 

(ii)

1 1 1 1 ln 27  ln 16 ln 5 ln(27) 3  ln(16) 4  ln 5  ln 3  ln 2  ln 5 3 4 6 3  2   ln   ln     5  5 

(i)

Using log rules:





 x  y ( x  y)   x2  y2   ln  ln x 2  y 2  ln(x  y )  ln   ln x  y    (x  y )   x  y  





Starting with the 2 ln x 1term:

(ii)

 

2 ln x 1  ln x 1

2

 1 1 2 2  ln x      ln x   ln 2   x 

Re-writing the original expression with logarithm rules:





 3 x3  5 x 2  1   1  3 2   2   ln 1  0 ln 3x  5x  ln  2   ln 3x  5  ln  x   3 x 5  x 





(iii)

2

e 2ln x  eln x  x 2

using logarithm rules

or

    x 2

e2ln x  eln x

(iv)

2

x

2

using power rules

The expression may not need to be simplified further, however it sheds some light on the relationship between natural logarithms and base 10 logarithms. This expression can be evaluated, numerically: 1 1 1    2.302585 log 10 e log 10 2.71828 ... 0.434294

Also worth consideration is the following: let y  log10 e raising each side to the power of 10 gives: e

10 y  10log10  e

taking the natural logarithm of each side: y ln 10  ln e  1 y ln 10  1

y

1 ln 10

and since y  log10 e , it follows that:

log 10 e 

3.

(i)

1 ln 10

or

ln 10 

1 log 10 e

Start with the application of Pythagorus’ theorem, to calculate the hypotenuse of this right-angled triangle:

a 2  b2  c 2 c  a 2  b 2  32  42  25  5 then, the sine and cosine of each angle can be calculated:

(ii)

sin  

opposite 3  hypotenuse 5

cos  

adjacent 4  hypotenuse 5

sin  

4 5

cos  

3 5

Angles can be calculated from any relevant trigonometric identity:

3   sin 1    36.87

5  4   sin 1    53.13 5 

(iii)

360  2π radians

  36.87  0.6435 radians   53 .13  0.9273 radians

(iv)

For any angle, A: 2 2 sin 2 A  sin A and cos 2 A  cos A 

2

9 3 2 sin      25 5 2

16 4 cos      25 5 2

cos 2   sin 2  

16 9  1 25 25

cos 2   sin 2   1

is a useful trigonometric identity that holds for all angles....