Hi All. Just wondering why answer to (i) is not 4.9e+4. an why answer to (iii) is not (9x)/2 or 4.5 x. Many thanks. Brian

% (i) Find f(1.2) in the form 'a x 10^n' where a is correct to 1 decimal place
% (ii) Find x for which g(x) = 3.5
% (iii) Find g(f(x)) in simplest form

clear, clc

syms x

f = exp(9*x); g = log(sqrt(x));

% (i)

f_at_x = subs(f,x,1.2); % evaluates f(x = 1.2) as symbolic
f_at_x = double(f_at_x); % converts to double (real)
f_at_x = round(f_at_x,1) % ??? round to 1 decimal place

f_at_x = 4.9021e+04

% (ii)

solve (g == 3.5, x)

ans = 

% (iii)

g_f_x = (subs(g,x,f));
g_f_x = simplify(g_f_x)

g_f_x = 

Answer 1

[Aquatris]

Your value in (i) for f(1.2) is 49020.80113638172, so when you round to 1 decimal, it becomes 49020.8 which in engineering notation is 4.9021e4 because of rounding the right most digit.

You should reread the question to understand why you are not getting 4.9 as answer (hint:question is asking a where f(1.2) = a * 10^n ) :)

syms x
f = exp(9*x); g = log(sqrt(x));
% (i)
f_at_x = subs(f,x,1.2); % evaluates f(x = 1.2) as symbolic
format long % change displasy setting
f_at_x = double(f_at_x) % converts to double (real)

f_at_x = 
     4.902080113638172e+04

f_at_x_rounded = round(f_at_x,1) % ??? round to 1 decimal place

f_at_x_rounded = 
     4.902080000000000e+04

format bank % change displasy setting
f_at_x_rounded

f_at_x_rounded = 
      49020.80

For part ii, log(x^n) = n*log(x) so both expressions are equivalent

fun1 = @(x) 0.5*log(exp(9*x));
fun2 = @(x) log(exp(4.5*x));
fun1(1.2) == fun2(1.2) % are they equal?

ans = logical
   1

[Brian Smyth]

Thanks for that Aquatris. But f_at_x_rounded, after 'bank format' , displays as 49020.80 . How do I finish by displaying as 4.9e+4? Thanks v much.

[Aquatris]

Now this is a different question. Now you are asking how to convert 49020.80113638172 to 4.9 * 10^4. Here is one way

x = 49020.80113638172; % represent as x = a * 10^n
n = floor(log10(x))
tmp_ = 1; % decimal accuracy for a, 1,2,3,4 etc
a = floor(x/10^(n-tmp_))/10^tmp_

a = 4.9000

newX = a*10^n

newX = 49000

[Voss]

x = 49020.80113638172;

fprintf('%.1e',x)

4.9e+04

[Brian Smyth]

Hi Voss. Super! Thanks very much for that. B

[Voss]

@Brian Smyth You're welcome!

Note that fprintf is for outputting formatted text to the command line or to text files. If instead you want to display formatted text in a plot, say, you could use sprintf, which supports the same options for formatting that fprintf does, but returns a character vector containing the formatted text:

x = 49020.80113638172;
plot(1:10)
str = sprintf('%.1e',x) % formatted text (character vector, str)

str = '4.9e+04'

text(5,8,str) % create a text object at point (5,8) showing that formatted text

[Brian Smyth]

Hi Voss.

That's good to know. Never quite understood the difference between the two (i..e fprintf and sprintf)! Thank you.

[Voss]

You're welcome!

Answer 2

[Torsten]

For (iii), use

syms x real

instead of

syms x 

[Brian Smyth]

Hi Torsten. Brilliant. Thanks a mil. Brian