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I am new to logarithms, and I need to find out the value this expression.

$ \ Given, \\x=\sqrt{log_{11}7}\\y=\sqrt{log_711} \\find : e^{y \ln{7} -x \ln{11}} $

Fghj
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3 Answers3

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considering only the Exponent we have $$\sqrt{\frac{\ln(11)\ln(7)^2}{\ln(7)}}-\sqrt{\frac{\ln(7)\ln(11)^2}{\ln(11)}}=0$$ after cancelling. Therefore your term is $1$

Dr. Sonnhard Graubner
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Since $log_a b=\frac {log_c b} {lob_c a}$, using this formula we get $$e^{\sqrt {\frac {ln\, 11} {ln \,7}}\cdot ln \, 7 -\sqrt {\frac {ln\, 7} {ln \,11}}\cdot ln \, 11 }$$ I guess you can finish it from here.

TStancek
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  • The finished answer is below ☺ – Fghj Jun 30 '17 at 06:15
  • I wanted to give you just the piece of information I thought you were missing and not spoil the fun part afterwards :) – TStancek Jun 30 '17 at 09:44
  • Oh no! Was it like that? I missed the fun part afterwards I'm so sorry and will try not to spoil such fun parts in future! – Fghj Jul 07 '17 at 18:03
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Since $log_a b=\frac {log_c b} {lob_c a}$, using this formula we get $$e^{\sqrt {\frac {ln\, 11} {ln \,7}}\cdot ln \, 7 -\sqrt {\frac {ln\, 7} {ln \,11}}\cdot ln \, 11 }$$

considering only the Exponent we have $$\sqrt{\frac{\ln(11)\ln(7)^2}{\ln(7)}}-\sqrt{\frac{\ln(7)\ln(11)^2}{\ln(11)}}=0$$ after cancelling. Now the expression equals $\ e^0 \ \\ $ Therefore answer is $1$

Fghj
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