To determine

To find: - The quantity as a single logarithm.

Answer

log104+log10a−13log10(a+1)=log104a(a+1)1/3

Explanation

Given Data:- log104+log10a−13log10(a+1)

Formula used: - If b>1, the function f(x)=logbx is a one to one, continuous increasing function with domain (0,∞) and range R . x,y>0 and r is any real number, then

a) logbxy=logbx+logby

b) logbxy=logbx−logby

c) logb(xr)=rlogbx

Since logb(xr)=rlogbx. Therefore,

log104+log10a−13log10(a+1)=log104+log10a−log10(a+1)1/3

Now,logbxy=logbx+logby, so above expression can be written as: 

 log104+log10a−13log10(a+1)=log104+log10a−log10(a+1)1/3=log10(4a)−log10(a+1)1/3

Since, logbxy=logbx−logby. So,

log104+log10a−13log10(a+1)=log104+log10a−log10(a+1)1/3=log10(4a)−log10(a+1)1/3=log104a(a+1)1/3

Hence, log104+log10a−13log10(a+1) is expressed as log104a(a+1)1/3 in a single logarithm.

Conclusion: log104+log10a−13log10(a+1)=log104a(a+1)1/3