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805+ (Hard)|   Exponents|      
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Bunuel
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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 31 Mar 2020, 03:27
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50% (02:55) correct 50% (02:37) wrong based on 138 sessions

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If \((5.55 )^x = (0.555)^y = 1000\), then the value of \(\frac{1}{x} − \frac{1}{y}\) is

A. 3
B. 1
C. 2/3
D. 1/3
E. 1/6


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D
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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 31 Mar 2020, 03:45
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Bunuel
If \((5.55 )^x = (0.555)^y = 1000\), then the value of \(\frac{1}{x} − \frac{1}{y}\) is

A. 3
B. 1
C. 2/3
D. 1/3
E. 1/6
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Please find below the video solution

Answer: Option D

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Peddi
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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 31 Mar 2020, 03:41
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5.55=1000^1/x, 0.555=(1000)^1/y 5.55/0.555=1000^1/x/1000^1/y , 10=1000^1/x-1/y , 10= 10^3(1/x- 1/y) ....> 1/x-1/y= 1/3

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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 31 Mar 2020, 03:38
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Bunuel
If \((5.55 )^x = (0.555)^y = 1000\), then the value of \(\frac{1}{x} − \frac{1}{y}\) is

A. 3
B. 1
C. 2/3
D. 1/3
E. 1/6
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Given \((5.55)^x = (0.555)^y = 1000 =10^3\)
Applying log,
\(x*log(5.55) = y*log(0.555) = 3*log10 = 3\)
\(x*log(5.55) = 3\)
--> \(x = \frac{3}{log5.55}\)
\(\frac{1}{x} = \frac{log 5.55}{3}\)

\(y*log(0.555) = 3\)
\(y*log(\frac{5.55}{10}) = 3\)
\(y*log(5.55) – y*log10 = 3\)
\(y*log(5.55) – y = 3\)
\(y = \frac{3}{(log(5.55) -1)}\)
\(\frac{1}{y} = \frac{(log(5.55) -1)}{3}\)

So, \(\frac{1}{x} - \frac{1}{y} = \frac{1}{3}\)

Option D
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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 11 Jul 2024, 18:35
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Bunuel
If \((5.55 )^x = (0.555)^y = 1000\), then the value of \(\frac{1}{x} − \frac{1}{y}\) is

A. 3
B. 1
C. 2/3
D. 1/3
E. 1/6


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I got this right by estimating (and perhaps a little luck). I know 5^4 = 625 so X is likely slightly larger than 4 —e stimate 4.5.

I to make 0.555^y larger we know y has to be a negative value. I estimate y to be -4.5.

1/4.5 - (1/-4.5) = 1/4.5 + 1/4.5 = 2/4.5 or slightly less than 1/2.

D is the closest answer.
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If (5.55 )^x = (0.555)^y = 1000, then the value of 1/x 1/y is 04 Sep 2025, 20:56
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