IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

Which is greater - $\sin 1 \ \text{or} \ \log_3 \sqrt 7$

Note: that is 1 radian not 1 degree.

gaurav_cheetah

sin 1 is o.84...
log 3(root) 7 is 0.88....

hence, the log one is greater

hsbhatt

obviously i am looking for a more decisive solution.

Anyone?

hsbhatt

there used to be a time in this forum where there would at least be a lively discussion on this forum if the question itself had not been demolished by now.

Whats up with you guys???

GoNik

sir actually everyones busy wid counselling...mayb thats why theres noone...sry

Decoder

let f(x) = sinx ....g(x)= lnx...

the latter is monotinic incresing...

let any sign( <=> ) ..taking it as " o" exist between ...for some x...

f(1) o g(7) / g(9) ...

subtracting 1...

f(1) - 1 o g(7) - g(9) / g(9)

multiplying -1....

so sign in o is changed...

now dividing by 2 both sides...

1 - f(1) / 2 o* g(9) -g(7) / 2g(9)...

now rhs is g'(x) ..for some 7
1 - f(1) / 2 o* g'(c) / g(9)...

my trial over...damn..

Conjurer

Maybe you can give us a hint ? The two terms seem so much unrelated? What should we look for in such problems?

Decoder

can we actually make something out of wic i have calculated in my failed trial...

please give us hint..wic topic is applied here...

hsbhatt

For a start 1 radian is closest to which familiar angle

feynmann

1 rad< 60 deg

again sine is an increasing fn in ( 0 to pi/2)

so sin 1< sqrt(3)/2

so we have to prove 7>3^(sqrt 3)

sandeepramesh

$\sin 1 < \sin 60^{\circ} = \frac {\sqrt{3}}{2}$

TPT $\sin 1 < \log_{3}{\sqrt{7}} \leftrightarrow \frac {\sqrt{3}}{2} < \log_{3}{\sqrt{7}} \implies \text{TPT} \sqrt{3} < \log_{3} {7} \implies \text{TPT} 3^{\sqrt{3}} < 7 \implies \text{TPT} 3^{\sqrt{3}-1} < 2.33 < \frac {7}{3} \implies \text{TPT} 3^{\sqrt{3}-1} < 3^{\frac {3}{4}} < 2.33 \implies \text{TPT} 3^3 < (2.33)^4$ which is true Smile

Hence proved Mr. Green Razz

sandeepramesh

Sorry Feynmann, didnt see your start :)

PS: Ignore the extra TPT at the bottom!

hsbhatt

My solution:

$\sin 1 < \sin 60^{\circ} = \frac{\sqrt 3}{2}}$

So, we have to prove that $3^{\sqrt 3} < 7$

After coming to this stage, you must still conclusively prove the inequality.

First I establish that $\sqrt 3 < \frac{7}{4}$ , this can easily be proved by squaring both sides etc.

Now if $3^{ \frac{7}{4}} < 7$ or equivalently 3^7 < 7^4 we are done.

3^7 = 2187; 7^4 = 2401 and so the inequality holds true.

sandeepramesh

To make it more simpler

Continuing my solution Mr. Green

$\text{TPT} 3^3 < (2.33)^4$

Now $(2.33)^4 > (2.3)^4 = (5.29)^2 > (5.25)^2 = \left(\frac {21}{4}\right)^2 = \frac {441}{16} = 27.5\cdots > 27 = 3^3$

Hence Proved

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