Question 1192142
<pre>
{{{2+expr(1/2)*log(5,(x-1))-3*log(5,(2z))+expr(2/3)*log(5,(y))-log(5,(a))}}}

The first term " 2 " must be written as a logarithm with base 5.  
Here is how to do that:

Let N be such that log<sub>5</sub>(N)</sub> = 2

Then by the definition of lagarithm, 5<sup>2</sup> = N,

so we write 2 as log<sub>5</sub>(5<sup>2</sup>) 

Now we have:

{{{log(5,5^2)+expr(1/2)*log(5,(x-1))-3*log(5,(2z))+expr(2/3)*log(5,(y))-log(5,(a))}}} 

We move all coefficients of logs as exponents of whatever the log is
being taken of:


{{{matrix(2,1,"",

log(5,5^2)+log(5,(x-1)^("1/2"))-log(5,(2z)^3)+log(5,(y)^("2/3"))-log(5,(a)))}}} 

Write all the terms that have " + " before them together and those
that have " - " before them together:

{{{matrix(2,1,"",

log(5,5^2)+log(5,(x-1)^("1/2"))+log(5,(y)^("2/3"))-log(5,(2z)^3)-log(5,(a))))}}}  

Group the terms with + and group those with - preceded by a minus sign:

{{{matrix(2,1,"",

(log(5,5^2)+log(5,(x-1)^("1/2"))+log(5,y^("2/3")))-(log(5,(2z)^3)+log(5,(a)))))}}}

Inside both parentheses, use the rule of logs that says
"the sum of logs is the log of the product"

{{{log(5,(5^2(x-1)^("1/2")y^("2/3")))-log(5,((2z)^3a)))}}}

Finally, use the rule of logs that says
"the difference of two logs is the log of the quotient (division)

{{{log(5,((5^2(x-1)^("1/2")y^("2/3"))/((2z)^3a))))}}}

You can finish by :

1. writing 2<sup>5</sup> as 25.
2. writing (2z)<sup>3</sup> as 2<sup>3</sup>z<sup>3</sup> and again as 8z<sup>3</sup>

{{{log(5,((25(x-1)^("1/2")y^("2/3"))/(8z^3a))))}}}

Edwin</pre>