SeekerOfKnowledge
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#1
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#1
Ok i even have the mark scheme for this and i still dont understand whats going on

4.
Given that 0<x<4

log5 (4-x) -2log5 x = 1

Find the value of x


the mark scheme doesnt show the steps, its so confusing.

Help me out here?
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Mathematician!
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(Original post by SeekerOfKnowledge)
Ok i even have the mark scheme for this and i still dont understand whats going on

4.
Given that 0<x<4

log5 (4-x) -2log5 x = 1

Find the value of x


the mark scheme doesnt show the steps, its so confusing.

Help me out here?
Use two rules here. \log a^b = b \log a and \log a - \log b = \log\frac{a}{b} . Also remember that 5^y is the inverse of  \log_5 y
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Pheylan
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(Original post by SeekerOfKnowledge)
log5 (4-x) -2log5 x = 1

Find the value of x
log_5(4 - x) - 2log_5x = 1

log_5(4 - x) - log_5x^2 = 1

log_5(\frac {4 - x}{x^2}) = 1

\frac {4 - x}{x^2} = 5

4 - x = 5x^2

5x^2 + x - 4 = 0

x = \frac {-1 \pm 9}{10}

x = \frac {4}{5}
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Mathematician!
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#4
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(Original post by Pheylan)
log_5(4 - x) - 2log_5x = 1

log_5(4 - x) - log_5x^2 = 1

log_5(\frac {4 - x}{x^2}) = 1

\frac {4 - x}{x^2} = 5

4 - x = 5x^2

5x^2 + x - 4 = 0

x = \frac {-1 \pm 9}{10}

x = \frac {4}{5}
...Use spoiler tags if you are going to put the full working. I could have done that, but what's the point? Did the OP learn anything? They may as well have inputted the question into Wolframalpha!
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vdub
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Is that the only answer?
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learnerinaction
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#6
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hi, can u help me with something? why did the logs suddenly disappear i dont get that.
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Sir Cumference
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#7
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#7
(Original post by learnerinaction)
hi, can u help me with something? why did the logs suddenly disappear i dont get that.
Please explain fully what you mean - why did the logs disappear from what?

It would be better if you started a new thread with your question since this one is 7 years old.
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