log3a+log9a=3/4. Find the exact value of a? Can somebody try to get a TO ROOT 3.
It is really difficult when we have to
guess what you mean! To what
base are these logarithms?
What you
wrote could be interpreted as \(\displaystyle log_{10}(3a)+ log{10}(9a)= 3/4\), the "common logarithms" since, in elementary courses base 10 is often
assumed. But that is the same as \(\displaystyle log_{10}(27a^2)= 3/4\) but a is NOT \(\displaystyle sqrt{3}\).
Another interpretation is \(\displaystyle log_e(3a)+ log_e(9a)= 3/4\),, the "natural logarithm". In more advanced courses, that is the logarithm that is typically use but, again, that would not give \(\displaystyle a= \sqrt{3}\)
It turns out, you mean the "3" and "9" as bases! Those
should be written as \(\displaystyle log_3(a)\) and \(\displaystyle log_9(a)\) or "log_3(a)" and log_9(a) if you don't want to use Latex. In any case, the base is a
subscript and the argument of the function is in parentheses.
So \(\displaystyle log_3(a)+ log_9(a)= 3/4\). If \(\displaystyle y= log_9(a)\), then \(\displaystyle a= 9^y= (3^2)^y= 3^{2y}\) so \(\displaystyle log_3(a)= 2y= 2log_9(a)\). The equation is \(\displaystyle 2log_9(a)+ log_9(a)= 3log_9(a)= log_9a^3= 3/4\). Can you solve that?
