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Logs

Write these as a single logarithm:-

c) 5log⁡aa+(13)log⁡a27+log⁡a25\log_aa + \left(\frac{1}{3}\right)\log_a27+ \log_a2

The book has provided the solution of log⁡a6a5\log_a6a^5

But I'm uncertain as to whether my solution is also valid:-

5log⁡aa+(13)log⁡a27+log⁡a2=log⁡aa5+loga3+loga2=log⁡a6+55\log_aa + \left(\frac{1}{3}\right)\log_a27+ \log_a2 = \log_aa^5 + log_a3 + log_a2 = \log_a6 + 5

Many thanks.
(edited 8 years ago)
Original post by supreme
Write these as a single logarithm:-

c) 5log⁡aa+(13)log⁡a27+log⁡a25\log_aa + \left(\frac{1}{3}\right)\log_a27+ \log_a2

The book has provided the solution of log⁡a6a5\log_a6a^5

But I'm uncertain as to whether my solution is also valid:-

5log⁡aa+(13)log⁡a27+log⁡a2=log⁡aa5+loga3+loga2=log⁡a6+55\log_aa + \left(\frac{1}{3}\right)\log_a27+ \log_a2 = \log_aa^5 + log_a3 + log_a2 = \log_a6 + 5

Many thanks.


It's correct; but it's not a single logarithm. So false.
Reply 2
Avatar for Jamdroid
Jamdroid
2
A single logarithm means just one term, with Log in it.

[Let]

First, use the power rule.
xLogy = Logy^x

So:
5Loga = Loga^5
(1/3)Log27 = Log27^(1/3) = Log3

You now have:
Loga^5 + Log3 + Log2

Now, use the product rule
Logx + Logy = Logxy

So:
Log(a^5x3x2)

You now have:
Log(6a^5)
Reply 3
Avatar for supreme
supreme
OP
4
Original post by Jamdroid
A single logarithm means just one term, with Log in it.

[Let]

First, use the power rule.
xLogy = Logy^x

So:
5Loga = Loga^5
(1/3)Log27 = Log27^(1/3) = Log3

You now have:
Loga^5 + Log3 + Log2

Now, use the product rule
Logx + Logy = Logxy

So:
Log(a^5x3x2)

You now have:
Log(6a^5)


Thanks :smile:
Reply 4
Avatar for supreme
supreme
OP
4
Original post by multiratiunculae
It's correct; but it's not a single logarithm. So false.


Yes, just shows I didn't read the question properly.

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