+0

# I'm not actually sure how to type this out, so I took a screenshot. Does anyone know how to solve this? http://prntscr.com/5a6cz7

0
1008
2

I'm not actually sure how to type this out, so I took a screenshot. Does anyone know how to solve this? http://prntscr.com/5a6cz7

Nov 26, 2014

#2
+114493
+10

Good answer anon although not many calcs do base three so you cannot just plug the numbers in and get an answer.

I am going to write the question sightly differently

$$\\If\;\;\;y=log_{3}3^6\\ then\\ 3^6={3}^y\\\\ Now it is very easy to see that y=6\\ so\\ log_33^6=6\\\\\\ Remember: \quad \boxed{If \quad y=log_ba\quad then \quad a=b^y}\\\\  You say this as \qquad y=log a base b\\\\ Also remember : \boxed{\mbox{A log is a power}}$$

Nov 26, 2014

#1
+5

The log to base 3 of any number is the power numeral when you express that number as a power of 3.

For example, if we write 9 as $${{\mathtt{3}}}^{{\mathtt{2}}}$$ then you see the power there is 2, so when we take logs to base 3,

we write log 9 = 2

Summarising, log $${{\mathtt{3}}}^{{\mathtt{2}}}$$ = 2

so log $${{\mathtt{3}}}^{{\mathtt{6}}}$$ = ......

If ever you're not sure your answer s right, you can always check using a calculator!

Nov 26, 2014
#2
+114493
+10
$$\\If\;\;\;y=log_{3}3^6\\ then\\ 3^6={3}^y\\\\ Now it is very easy to see that y=6\\ so\\ log_33^6=6\\\\\\ Remember: \quad \boxed{If \quad y=log_ba\quad then \quad a=b^y}\\\\  You say this as \qquad y=log a base b\\\\ Also remember : \boxed{\mbox{A log is a power}}$$