Question 253217
Not quite!
If {{{f(x) = Log[2](x)}}} find {{{f(16)}}} 
Recall that the definition of the logarithm of a number is the "power" to which the "base" must be raised to equal that number. Or {{{Log[b](x) = y}}} which can be expressed as: {{{b^y = x}}}
In this problem, the 'base' is given as 2, and to find f(16) you must replace the x in the f(x) of the problem with 16, so now you have:
{{{f(16) = Log[2](16)}}}
Let's rewrite the problem a bit replacing {{{f(16)}}} wth {{{y}}}:
{{{Log[2](16) = y}}}, then according to the above definition, we can write this as ({{{Log[b](x) = y}}}--->{{{b^y = x}}}:
{{{2^y = 16}}} Substitute {{{16 = 2^4}}}
{{{2^y = 2^4}}} Now since the bases are equal, the exponents must be equal, so...
{{{y = 4}}} but {{{y = f(16)}}} so...
{{{highlight(f(16) = 4)}}}