When Is This Logarithm Inequality True?

Algebra Level 1

True or False?

For all positive integers $n \geq 2$ ,

$\log_{10} (n-1) + \log_{10} (n+1) < 2 \log_{10} n.$

True only if

n \leq 10

False always True only if

n \leq 100

True always

1 solution

Yashwanth Manivannan
Mar 5, 2016

Using $\log _{ x }{ m } +\quad \log _{ x }{ n } =\log _{ x }{ mn } \\$

We can arrive at,

$\log _{ 10 }{ (n-1) } +\log _{ 10 }{ (n+1) } \quad <\quad 2\log _{ 10 }{ \quad n } \\ \log _{ 10 }{ ((n-1)(n+1)) } \quad \quad \quad \quad <\quad \log _{ 10 }{ \quad { n }^{ 2 } } \quad \quad [x\quad \log _{ a }{ b } \quad =\quad \log _{ a }{ { b }^{ x } } ]\\ \log _{ 10 }{ ({ n }^{ 2 } } -1)\quad \qquad \qquad \qquad <\quad \log _{ 10 }{ { n }^{ 2 } }$

Now,

${ n }^{ 2 }-1\quad <\quad { n }^{ 2 }$

This condition is always true.

why is 1 not included because it also satisfy the inequality

Moleseng Motsumi - 2 years ago

For $n=1$ , the argument in the first logarithm i.e. $n-1$ , becomes zero, which is not permitted. Additionally, for $n=1$ , the logarithm on the RHS becomes zero, making the statement a false one (which was already undefined in the first place).

Krish Shah - 1 year, 2 months ago

Domain : n > 1 thus solution set is (1, infinity)

Rakesh Ahuja - 8 months, 4 weeks ago

@Moleseng Motsumi, 1 indeed satisfies the inequality but we have to ignore it, because we have already been given an arbitary domain , n>(=) 2

Uday Bhullar - 6 months ago

When Is This Logarithm Inequality True?

1 solution

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