Lesson derivation of change of base rule for logarithms

Source code of 'derivation of change of base rule for logarithms'

This Lesson (derivation of change of base rule for logarithms) was created by by gonzo(654)

: View Source, Show
About gonzo:

-----
The rule for changing bases in logarithms is as follows:
{{{log(a,x) = log(b,x)/log(b,a)}}}
-----
the basic rules for logarithms states:
{{{log(a,x) = y}}}
if and only if:
{{{a^y = x}}}
-----
{{{log(a,x) = y}}} is the logarithmic form of the equation.
{{{a^y = x}}} is the exponential form of the equation.
-----
since for all real numbers, we should always be able to find a "b" and a "c" such that:
{{{a=b^c}}}
then:
{{{a^y = x}}}
becomes:
{{{(b^c)^y = x}}}
which becomes:
{{{b^(c*y) = x}}}
-----
the original formula of:
{{{a^y = x}}}
has been transformed to:
{{{b^(c*y) = x}}}
because:
{{{a=b^c}}}
-----
by the basic rule for logarithms:
{{{b^(c*y) = x}}}
if and only if:
{{{log(b,x) = c*y}}}
dividing both sides of this formula by c and it becomes:
{{{(log(b,x))/c = y}}}
-----
we can solve for c as follows:
-----
since {{{a=b^c}}} is true if and only if:
{{{log(b,a) = c}}}
then:
we can substitute {{{log(b,a)}}} for {{{c}}}.
-----
the formula:
{{{(log(b,x))/c = y}}}
becomes:
{{{log(b,x)/log(b,a) = y}}}
-----
since we started with:
{{{log(a,x) = y}}}
and we derived:
{{{log(b,x)/log(b,a) = y}}}
then:
{{{log(a,x) = log(b,x)/log(b,a)}}}
because they are both equal to y.
-----
the change of base formula for logarithms has been derived and it all stems from the fact that any base can be represented by any other base raised to an exponent (a = b^c).
-----
i could not find ways to prove this in actual practice other then through the use of the calculator.
there are two bases i could work with (10 and e).  the base of e is the ln function in the calculator.  the base of 10 is the log function in the calculator.
-----
i'll do 1 example to show you how this works:
-----
15^3 = 3375
if and only if {{{log(15,3375) = 3)}}}
since i can't solve this directly, i'll use the change of base formula with the aid of the ln function of the calculator.
{{{log(15,3375) = 3)}}} = {{{log(e,3375)/log(e,15)}}} = 3.
-----
i'll solve it again using the change of base formula with the aid of the log function of the calculator.
{{{log(15,3375) = 3)}}} = {{{log(10,3375)/log(10,15)}}} = 3.
-----