# Log comparision question

• Jan 14th 2009, 05:03 PM
Uncle6
Log comparision question
"Which is greater, \$\displaystyle \log_3{5} or \log_5{11}\$? Explain"

Although u can plug and chug into your calculator, this question seems like it's meant to be solved analytically.

I got to:

\$\displaystyle \log_3{5}=a , \log_5{11}=b\$
\$\displaystyle 3^a=5, 5^b=11\$
\$\displaystyle (3^a)^b=11
\$

Dunno what's next. I can figure out that a<2 and ab>2
• Jan 14th 2009, 05:21 PM
Jhevon
Quote:

Originally Posted by Uncle6
"Which is greater, \$\displaystyle \log_3{5} or \log_5{11}\$? Explain"

Although u can plug and chug into your calculator, this question seems like it's meant to be solved analytically.

I got to:

\$\displaystyle \log_3{5}=a , \log_5{11}=b\$
\$\displaystyle 3^a=5, 5^b=11\$
\$\displaystyle (3^a)^b=11
\$

Dunno what's next. I can figure out that a<2 and ab>2

think of it this way:

we have \$\displaystyle 3^a = 5\$ and \$\displaystyle 5^b = 11\$

Note that 5 is less than twice 3, while 11 is more than twice 5. in other words, \$\displaystyle b\$ has the task of increasing the base 5 a larger magnitude (relatively speaking) than \$\displaystyle a\$ has to do for 3. \$\displaystyle b\$ is larger

this is a more intuitive approach. which i suppose is the level of thinking this problem requires, since it is elementary/middle school math. there are more rigorous proofs, i'm sure
• Jan 15th 2009, 02:50 PM
Uncle6
Thanks, but it still is quite difficult to understand.
• Jan 15th 2009, 03:49 PM
Jhevon
Quote:

Originally Posted by Uncle6
Thanks, but it still is quite difficult to understand.