How do you compare logs ?

Algebra Level 2

log ⁡ 3 108 OR log ⁡ 5 375 \Large \color{#20A900}{\log_3108} \quad \color{#D61F06}{\text{OR}} \quad \color{#3D99F6}{\log_5375}

Which one of the numbers above is greater?

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Both are equal in value \color{#D61F06}{\text{Both are equal in value}} log ⁡ 3 108 \color{#20A900}{\log_3108} log ⁡ 5 375 \color{#3D99F6}{\log_5375}

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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8 solutions

Abhishek Sharma
May 12, 2015

We need to compare log ⁡ 3 108 \log _{ 3 }{ 108 } and log ⁡ 5 375 \log _{ 5 }{ 375} .

We have,

log ⁡ 3 108 \log _{ 3 }{ 108 } = log ⁡ 3 ( 27 × 4 ) \log _{ 3 }{ (27 \times 4 } ) = log ⁡ 3 ( 3 3 × 4 ) \log _{ 3 }{( {3}^{3} \times 4 )} = log ⁡ 3 3 3 + log ⁡ 3 4 \log _{ 3 }{ {3}^{3}} + \log_{3}{4} = 3 + log ⁡ 3 4 3+\log_{3}{4}

Similarly,

log ⁡ 5 375 \log _{ 5 }{ 375 } = log ⁡ 5 ( 125 × 3 ) \log _{ 5 }{ (125 \times 3 }) = log ⁡ 5 ( 5 3 × 3 ) \log _{ 5 }{ ({5}^{3} \times 3 )} = log ⁡ 5 5 3 + log ⁡ 5 3 \log _{ 5 }{ {5}^{3}}+\log_{5}{3} = 3 + log ⁡ 5 3 3+\log_{5}{3}

Now consider,

log ⁡ 3 4 > log ⁡ 5 3 \log_{3}{4}>\log_{5}{3}

Adding 3 3 on both sides,

3 + log ⁡ 3 4 > 3 + log ⁡ 5 3 3+\log_{3}{4}>3+\log_{5}{3}

Using the already proved results we have,

log ⁡ 3 108 > log ⁡ 5 375 \log _{ 3 }{ 108 }>\log _{ 5 }{ 375 }

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good! A simpler approach is to apply the inequalities for the pairs 108 , 81 108,81 and 375 , 625 375,625 . Do you see why?

log ⁡ 3 108 > log ⁡ 3 81 = 4 and log ⁡ 5 375 < log ⁡ 5 625 = 4 \log_3108\gt \log_381=4\quad\textrm{and}\quad \log_5375\lt \log_5625=4

∴ log ⁡ 5 375 < 4 < log ⁡ 3 108 \therefore\quad \log_5375\lt 4\lt \log_3108

This follows from the monotonicity of the logarithm function, or more specifically, the real-valued base 5 5 -logarithm is strictly increasing. More generally speaking, we used the following fact here:

0 < a < b ⟹ log ⁡ x a < log ⁡ x b ∀ x ∈ ( 1 , ∞ ) 0\lt a\lt b\implies \log_xa\lt \log_xb~~\forall~x\in(1,\infty)

Prasun Biswas - 6 years, 1 month ago

Nice solution, I did it in an almost identical way (by taking multiples of 3 and 5 out equally for both logs until I could compare them by observation).

Michael Fuller - 6 years, 1 month ago
Chew-Seong Cheong
May 13, 2015

log ⁡ 3 108 = log ⁡ 10 108 log ⁡ 10 3 = log ⁡ 10 4 × 3 3 log ⁡ 10 3 = 3 log ⁡ 10 3 + log ⁡ 10 4 log ⁡ 10 3 = 3 + log ⁡ 10 4 log ⁡ 10 3 > 4 log ⁡ 5 375 = log ⁡ 10 375 log ⁡ 10 5 = log ⁡ 10 3 × 5 3 log ⁡ 10 5 = 3 log ⁡ 10 5 + log ⁡ 10 3 log ⁡ 10 5 = 3 + log ⁡ 10 3 log ⁡ 10 5 < 4 \begin{aligned} \log_3{108} & = \dfrac{\log_{10}{108}}{\log_{10}{3}} = \dfrac{\log_{10}{4\times 3^3}}{\log_{10}{3}} = \dfrac{3\log_{10}{3}+\log_{10}{4}}{\log_{10}{3}} \\ & = 3+ \dfrac{\log_{10}{4}}{\log_{10}{3}} \color{#3D99F6}{> 4} \\ \log_5{375} & = \dfrac{\log_{10}{375}}{\log_{10}{5}} = \dfrac{\log_{10}{3\times 5^3}}{\log_{10}{5}} = \dfrac{3\log_{10}{5}+\log_{10}{3}}{\log_{10}{5}} \\ & = 3+ \dfrac{\log_{10}{3}}{\log_{10}{5}} \color{#3D99F6}{< 4} \end{aligned}

Therefore, log ⁡ 3 108 > log ⁡ 5 375 \boxed{ \log_3{108}} > \log_5{375}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Nicely done. Here's my approach:

log ⁡ 3 108 > log ⁡ 3 81 = log ⁡ 3 3 4 = 4 log ⁡ 3 3 = 4 \log_3 108 > \log_3 81 = \log_3 3^4 = 4 \log_3 3 = 4

= 1 + 3 log ⁡ 5 5 = 1 + log ⁡ 5 125 > log ⁡ 5 3 + log ⁡ 5 125 = log ⁡ 5 375 = 1 + 3 \log_5 5 = 1 + \log_5 125 > \log_5 3 + \log_5 125 = \log_5 375 .

Li Yuelin
May 11, 2015

l o g 3 108 = 3 l o g 3 4 log_{3}108 = 3log_{3}4

l o g 5 375 = 3 l o g 5 3 log_{5}375 = 3log_{5}3

l o g 3 4 > l o g 5 3 log_{3}4 > log_{5}3

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Wrong. Check your working again.

Hobart Pao
May 13, 2015

I looked at it this way: log ⁡ 3 81 = 4 \log_{3} 81 = 4 and log ⁡ 5 625 = 4 \log_{5} 625 = 4 . 625 is already much larger than 375, but with 81, there's a still a little ways to go until 108. So, log ⁡ 3 108 > log ⁡ 5 375 \log_{3} 108 > \log_{5} 375

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Devin Ky
May 13, 2015

log 3 108 > log 3 81 > 4

log 5 375 < log 5 625 < 4.

The green one is greater than 4 and the blue one is less than 4. Thus the result is obvious.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Careful with your notations, it should be: . . . log ⁡ 3 81 = 4 ... \log_3 81 = 4 .

l o g 3 108 = l o g 3 ( 27 ⋅ 4 ) = 3 + l o g 3 4 > 4 ) log_{3} 108 = log_{3} (27\cdot 4) = 3+log_{3}4 > 4) . l o g 5 375 = l o g 5 ( 3 ⋅ 125 ) = 3 + l o g 5 3 < 4 log_{5} 375 =log_{5} (3\cdot 125) = 3+log_{5} 3 < 4

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Raushan Sharma
May 13, 2015

Let log ⁡ 3 108 = x \log_3 108 = x Then, 3 x = 108 3^{x} = 108 So, 4 < x < 5 4 < x < 5 Let log ⁡ 5 375 = y \log_5 375 = y Then, 5 y = 375 5^{y} = 375 So, 3 < y < 4 3 < y < 4 So, y < 4 < x y < 4 < x Therefore, x > y x > y ⇒ \Rightarrow log ⁡ 3 108 \log_3 108 > > log ⁡ 5 375 \log_5 375

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Andrew Tawfeek
May 13, 2015

Remember:

log ⁡ 3 108 v s . log ⁡ 5 375 = log ⁡ 108 log ⁡ 3 v s . log ⁡ 375 log ⁡ 5 \quad \log _{ 3 }{ 108\quad vs.\quad \log _{ 5 }{ 375 } } \\ =\quad \frac { \log { 108 } }{ \log { 3 } } \quad vs.\quad \frac { \log { 375 } }{ \log { 5 } }

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

So? Making one expression equivalent to another does not necessarily make it closer to the answer.

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