$$\log\sqrt [ 3 ]{ \frac { x+2 }{ x^{ 4 }(x^{ 2 }+4) } } $$
How is this answer incorrect?
$$\frac { 1 }{ 3 } [\log(x+2)-(4\log x+\log(x^ 2+4))]$$
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3Your answer is absolutely correct. – symmetricuser Oct 02 '14 at 07:28
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The answer key saids otherwise. – Cherry_Developer Oct 02 '14 at 07:29
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What did the answer key say? – Taladris Oct 02 '14 at 07:30
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$$\frac { 1 }{ 3 } [log(x+2)-4logx-log(x^{ 2 }+4)]$$ – Cherry_Developer Oct 02 '14 at 07:31
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2If $x=-1$, or thereabouts, there is an issue. – André Nicolas Oct 02 '14 at 07:32
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2That's the same thing. They just distributed the $-$ sign over $4 \log x + \log(x^2+4)$. – symmetricuser Oct 02 '14 at 07:32
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It's very hard to see how the text books answer is the same thing once combined... – Cherry_Developer Oct 02 '14 at 07:33
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with this simplify you have to know that your domain will be changed and that is not exactly with your initial question. – Panda Oct 02 '14 at 07:35
3 Answers
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There is nothing wrong with your answer though the person asking the question might be expecting it to be expanded into $$\tfrac { 1 }{ 3 } \log(x+2)-\tfrac43\log x-\tfrac13\log(x^ 2+4)$$ to remove many of the brackets.
Henry
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with this simplify you have to know that your domain will be changed and that is not exactly with your initial question.suppose you want to simplify
this you can't input negative integer in this,but if you simplify that:
that you can input negative numbers,so domain changed.
Panda
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2This answer is incorrect. You can't put negative numbers into $\log(x)$ either. – rae306 Oct 02 '14 at 08:06
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A better answer would be that $\log(x^4)$ does not have the same domain as $4\log(x)$: consider $x\lt 0$. This affects this question for $-2 \lt x \lt 0$ – Henry Oct 02 '14 at 08:36
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Your answer is correct, with reservation about function domain. $Dom(log\sqrt[3]{\frac{x+2}{x^{4}(x^{2}+4)}})=\left(-2,\infty\right)/\{0\}$ $$ Dom[\frac{1}{3}(log(x+2)-(4log(x)+log(x^{2}+4)]=(0,\infty)$$ For example, if $x=-1$, original function give $-log(5)/3$, but your formula "Error"
Slepecky Mamut
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