Question 250423
the rules of logarithms state:


{{{log(a*b) = log(a) + log(b)}}} (rule 1)


{{{log(a/b) = log(a) - log(b)}}} (rule 2)


{{{log(a^b) = b * log(a)}}} (rule 3)


your expression is:


{{{4 * log(a,y) + (1/7) * log(a,z)}}}


{{{4 * log(a,y) = log(a,y^4)}}} by rule 3.


{{{(1/7) * log(a,z) = log(a,z^(1/7))}}} by rule 3.


your expression becomes:


{{{log(a,y^4) + log(a,z^(1/7))}}}


this becomes {{{log(a,(y^4*z^(1/7)))}}} by rule 1.


since {{{z^(1/7)}}} can also be written as {{{root(7,z)}}}, this can also be written as:


{{{log(a,(y^4*root(7,z)))}}}


you can prove that your original expression of {{{4 * log(a,y) + (1/7) * log(a,z)}}} is the same as your final expression of {{{log(a,(y^4*root(7,z)))}}} by substituting for the values of a, y, z.


I used a = 10 because this base is solvable by using the LOG function of the calculator.


I let y = 7
I let z = 15


I got the same answer for the original expression and the final expression which says that I made the translation correctly.