Question 286328
{{{log(c, (5)) = a}}}
{{{log(c, (4)) = b}}}
To find {{{log(c, (2))}}} we will need to use the properties on the base c logarithms we have. There are properties for products, quotients and powers of arguments. Since 4 is a power of 2 we can use {{{log(c, (4)) = b}}} and the property about powers in the argument of a logarithm to find {{{log(c, (2))}}}:
{{{log(c, (4)) = b}}}
{{{log(c, (2^2)) = b}}}
Using the property, {{{log(x, (p^q)) = q*log(x, (q))}}}, to move the exponent out of the argument:
{{{2log(c, (2)) = b}}}
Now we jsut divide by two:
{{{log(c, (2)) = b/2}}}<br>
We now have three base c logarithms. For {{{log(c, (10))}}} we look for a product, quotient or power of 2, 4 and/or 5 which makes 10. Obviously 2*5=10 so:
{{{log(c, (10))}}}
{{{log(c, (2*5))}}}
Now we can use the property for products in an argument, {{{log(x, (p*q)) = log(x, (p)) + log(x, (q))}}}, to separate the 2 and the 5:
{{{log(c, (2)) + log(c, (5))}}}
We can replace these logarithms with what we know them to be:
{{{b/2 + a}}}