anonymous
  • anonymous
can any one solve this?(log(a)(x^3))+(log(a)(sqrtx))= answer has to be in klog(a)x form thanks already :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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amistre64
  • amistre64
is that base a?
amistre64
  • amistre64
log[a](x^3) like this?
amistre64
  • amistre64
3loga(x) + (1/2)loga(x)

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More answers

amistre64
  • amistre64
loga(x)(3+(1/2)) ??
anonymous
  • anonymous
Amistre, I think contact has been lost lol.
amistre64
  • amistre64
looks right lol (6/2) loga (x)
amistre64
  • amistre64
:) but the math remains lol
anonymous
  • anonymous
Hahahaa, so very true!
amistre64
  • amistre64
(7/2) loga(x)
anonymous
  • anonymous
yes a is the base, soz my internet played up :S
amistre64
  • amistre64
can we factor out the loga(x) like that tho?
anonymous
  • anonymous
\[\log_{a} x^3+\log_{a} sqrtx\]
amistre64
  • amistre64
if we multiply it togeter again we get: loga[sqrt(x^7)] so its good
amistre64
  • amistre64
(7/2) loga(x) is your answer
anonymous
  • anonymous
where did 7 come from? soz but i dont get the stuff you lot did at the top :S
amistre64
  • amistre64
3 + 1/2 = 3' 1/2 to get it into an improper fraction (topheavy) we multiply 3 by the denom and add the numerator; 3(2) = 6 + 1 = 7.... 7/2
amistre64
  • amistre64
OR we could go the route of changing addition of logs back into multiplication of logs
anonymous
  • anonymous
ok :) multiplication logs?
amistre64
  • amistre64
loga(x^3 sqrt(x)) = loga(sqrt(x^7)) = loga(x^(7/2)) (7/2) loga(x)
amistre64
  • amistre64
loga(x) + loga(y) = loga(xy)
amistre64
  • amistre64
loga(x^n) = n loga(x) as well
anonymous
  • anonymous
ok so 7/2logax is the answer right?
amistre64
  • amistre64
yes; (7/2) loga(x) is the answer
anonymous
  • anonymous
\[7/2\log_{a} x\]
anonymous
  • anonymous
yay! ill give u all medals :) thx
amistre64
  • amistre64
im gonna need a new cigar box to keep these things in ;)
anonymous
  • anonymous
:p

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