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