Question 818767
Note: In the future please put parentheses around function arguments.<br>
{{{7log((16/15))+5log((25/24))+3log((81/80))=log((2)))}}}
What we will be doing is using properties of logarithms ({{{log(a, (p)) + log(a, (q)) = log(a, (p*q))}}}, {{{log(a, (p)) - log(a, (q)) = log(a, (p/q))}}} and/or {{{n*log(a, (p)) = log(a, (p^n))}}}) to condense the entire left side into a single logarithm (hopefully log(2)).<br>
We'll start by using the third property to move the coefficients of each log into its argument as its exponent:
{{{log(((16/15)^7))+log(((25/24)^5))+log(((81/80)^3))=log((2)))}}}
Next we can use the first property to combine the three logs:
{{{log((((16/15)^7)*((25/24)^5)*((81/80)^3))))=log((2)))}}}
To make simplifying the argument easier, I am going to rewrite each numerator and denominator in terms of prime factors:
{{{log((((2^4/(3*5))^7)*((5^2/(2^3*3))^5)*((3^4/(2^4*5))^3))))=log((2)))}}}
Now we can use the {{{(a/b)^n = a^n/b^n}}} rule to raise each of these fractions to a power:
{{{log(((2^28/(3^7*5^7))*(5^10/(2^15*3^5))*(3^12/(2^12*5^3))))=log((2)))}}}
Using the {{{a^n * a^m = a^(n+m)}}} rule to multiply the numerators and denominators:
{{{log(((2^28*5^10*3^12)/(3^12*5^10*2^27)))=log((2)))}}}
The 3's and 5's all cancel out. And we can use the {{{a^n/a^m = a^(n-m)}}} rule to divide the 2's:
{{{log((2^(28-27)))=log((2)))}}}
{{{log((2^1))=log((2)))}}}
{{{log((2))=log((2)))}}} Check!