Question 117426
1.)  For any polynomial equation with real coefficients, complex roots occur in conjugate pairs.
So, if 5i i.e. (0 + 5i) is a root then (0 - 5i) is the other root.
Hence, the 4 zeros are x = -1, 2, 5i, - 5i
Hence the polynomial is 
{{{(x - (-1))(x - 2)(x - 5i)(x - (-5i))}}}
= {{{(x + 1)(x - 2)(x - 5i)(x +5i)}}}
= {{{(x + 1)(x - 2)(x^2 - 25i^2)}}}
= {{{(x + 1)(x - 2)(x^2 + 25)}}}
= {{{(x^2 - x - 2)(x^2 + 25)}}}
= {{{x^4 - x^3 + 23x^2 - 25x - 50}}}

2.)  
f(x) = 5x + 1
5x = f(x) - 1
{{{x = (f(x)-1)/5}}}
Hence, {{{f^(-1)(x) = (x - 1)/5}}}


Verification:
{{{f(f^(-1)(x)) = 5f^(-1)(x) + 1 = 5(x-1)/5 + 1 = x}}}

3.a) {{{log(10,x+4) - log (10,2x) = 0}}}
{{{log(10,(x+4)/2x) = 0}}}
{{{(x+4)/2x = 1}}}
{{{(x+4) = 2x}}}
{{{x = 4}}}

3.b) {{{2log(e,x) + 5 = 7}}}
{{{2log(e,x) = 7 - 5 = 2}}}
{{{log(e,x) = 1}}}
{{{x = e}}}

4.) Try yourself!