Question 106845
Solve g(x) = 0 for x, if 
{{{g(x)=log((x))+8*log x+log((x^2))-log((x^3))-16}}}
The original equation is re-written to group like terms. 
Here are the logarithm rules for reference
{{{log((a))+log((b))=log ((ab))}}}
{{{log((a))-log((b))=log ((a/b))}}}
{{{k log((a)) = log((a^k))}}}
Now let's appply the rules to your equation, 
{{{g(x)=log((x))+8*log x+log((x^2))-log((x^3))-16}}}
{{{g(x)=9*log((x))+log((x^2))-log((x^3))-16}}}
{{{g(x)=log((x^9))+log((x^2))-log((x^3))-16}}}
{{{g(x)=log((x^9*(x^2)/x^3))-16}}}
{{{g(x)=log((x^(9+2-3)))-16}}}
{{{g(x)=log((x^(8)))-16}}}
When g(x)=0, then
{{{0=log((x^(8)))-16}}}
{{{log((x^(8)))=16}}}
{{{x^8=10^16}}}
{{{x^8=(10^2)^8}}}
{{{x=(10^2)}}}
{{{highlight(x=100)}}}
Check your answer
{{{g(x)=log((x))+8*log x+log((x^2))-log((x^3))-16}}}
{{{g(100)= log(100)+8*log(100)+log(100^2)-log(100^3)-16}}}
{{{0=2+8*2+4-6-16}}}
{{{0=0}}}
Good answer.