If $z= - 0.1887\cdot(x^{0.7637})\cdot(y^{0.2306})$
Its natural logarithm will be
$\ln(z) = - [ \ln(0.1887) + 0.7637 \ln(x) + 0.2306 \ln(y)]$
or
$\ln(z) = - [ \ln(0.1887) - 0.7637 \ln(x) - 0.2306 \ln(y)]$?
Thank you in advance
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neither of them – Berci Dec 04 '13 at 22:14
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@Berci.. Thankyou! – Syeda Dec 04 '13 at 22:15
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1Hint: To write multidigit exponents, put them in braces. So x^{0.7637} gives $x^{0.7637}$ To get functions in the right font, put a backslash first, so \ln gives $\ln$ – Ross Millikan Dec 04 '13 at 22:45
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@RossMillikan.. Done!! Thankyou.. :) – Syeda Dec 04 '13 at 22:48
2 Answers
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The logarithm of a negative number is not defined among real numbers.
(It can have infinitely many values among complex numbers.)
But, for example $x^A / \,y^B = A\ln x - B\ln y$ if you want subtraction, or $(1/s)^A \cdot x^B\cdot y^C = -A\ln s+B\ln x+C\ln y\ $ if you like.
Berci
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@Berci.. okay! Thanks.. Actually I have expression in this form only! I don't know how to find its natural log!! – Syeda Dec 04 '13 at 22:19
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@Berci..How? Can you suggest it for my Question.. My both answer seems to be incorrect.. – Syeda Dec 04 '13 at 22:29
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If an answer is required, maybe try $$ \ln(z) = \ln(-1) + \ln(0.1887) + 0.7637 \ln(x) + 0.2306 \ln(y) $$ As others stated, this is not a real number. But maybe some complex number answer is intended?
GEdgar
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