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Sir que. No. 5.. can u help me... and if u hav some easy tricks to do this que.

Sir que. No. 5.. can u help me... and if u hav some easy tricks to do this que.

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Grade:12

3 Answers

Ajay
209 Points
7 years ago
For these type of problems you should convert all logs to a common base e.  Also you should know graph of log function so that you can determine if Log a is greater or less than log b.
Given\quad x=\quad { log }_{ 3 }5\quad and\quad y\quad =\quad { log }_{ 17 }25\\ To\quad change\quad the\quad base\quad we\quad use\quad formula\quad { log }_{ a }x\quad =\quad \frac { { log }_{ b }x }{ { log }_{ b }a } .\\ Changing\quad base\quad to\quad e\quad for\quad both\quad x\quad and\quad y\quad we\quad get\\ x\quad =\quad \frac { { log }_{ e }5 }{ { log }_{ e }3 } \quad or\quad { log }_{ e }5\quad =\quad x{ log }_{ e }3\quad --------(1)\\ y\quad =\quad \frac { { log }_{ e }25 }{ { log }_{ e }17 } \quad =\quad \frac { 2{ log }_{ e }5 }{ { log }_{ e }17 } \quad ------(2)\\ from\quad eq\quad 1\quad substitue\quad { log }_{ e }5\quad in\quad eq\quad 2\\ y=\quad \quad \frac { 2x{ log }_{ e }3 }{ { log }_{ e }17 } \\ y/x\quad =\quad \frac { 2{ log }_{ e }3 }{ { log }_{ e }17 } \quad =\quad \frac { { log }_{ e }9 }{ { log }_{ e }17 }.
To be continued
Ajay
209 Points
7 years ago
Here is rest of solution …................................................................................
Now\quad we\quad have\quad both\quad logs\quad in\quad same\quad base\quad its\quad easy\quad to\quad compare\quad y\quad and\quad x.\quad \\ Since\quad log\quad to\quad base\quad e\quad is\quad increaing\quad function\quad we\quad have\\ { log }_{ e }9\quad <\quad \quad { log }_{ e }17\quad or\quad \frac { { log }_{ e }9 }{ { log }_{ e }17 } \quad <\quad 1.\\ or\quad y/x\quad <\quad 1\\ or\quad y<\quad x
mycroft holmes
272 Points
7 years ago
Shorter solution:
 
1/x = log5 3  and 1/y = log25 17 = log5\sqrt{17}} >log53 
 
i.e. 1/x 0 we have x>y

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