Question 116270
Given:
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{{{log(3,root(4,(x^5*y^4)/81))}}}
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Your work was:
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{{{log(3,((x^5*y^4)^(1/4)))-log(3,((81)^(1/4)))}}} <=== ok
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{{{log(3, (x^(5/4))*( y^(4/4))) - (1/4)*log(3,81)}}}<=== ok
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{{{(5/4)* log(3,x) + (4/4)* log(3,y) - 9/4}}} <=== mistake in last term
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You have a mistake in evaluating {{{(-1/4) *log(3,81)}}}. Replace 81 with {{{3^4)}}} and the term
becomes:
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{{{(-1/4) *log(3,3^4)}}}
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Bring the exponent 4 out as a multiplier and the result is:
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{{{(-1/4)*4 *log(3,3)}}}
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But {{{log(3,3) = 1}}}
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so this reduces to:
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{{{(-1/4) * 4 * 1 = -1}}}
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So in your last equation replace - 9/4 with -1 and you have:
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{{{(5/4)* log(3,x) + (4/4)* log(3,y) - 1}}}
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and in the second term note that {{{4/4 = 1}}} so the expression reduces to:
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{{{(5/4)* log(3,x) + log(3,y) - 1}}}
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That should be enough to convince your teacher that you know what you are doing. I'm not sure
what else could be productively done on this problem. If you feel that it needs more work,
post it again with the above change to your work and see if some other tutor has another idea
about how to go further.
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Hope this helps you and the study group. Check my work and ensure that I haven't made a "late night"
error.
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