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log help needed

write 3log5-log2 as a single logarithm

do i do 5log3-log2
so 15log-log2

so log7.5? is this right

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Reply 1
Avatar for TGH1
TGH1
6
Original post by dongonaeatu
write 3log5-log2 as a single logarithm

do i do 5log3-log2
so 15log-log2

so log7.5? is this right


No, as you haven't applied the log rules correctly.

alnb = ln(b^a)
lnc - lnd = ln(c/d)
Reply 2
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
write 3log5-log2 as a single logarithm

do i do 5log3-log2
so 15log-log2

so log7.5? is this right


Remember, alogb=logba and logb−loga=log(ba) \displaystyle alogb=logb^a \ and \ logb-loga=log\left(\frac{b}{a}\right)
Reply 3
Avatar for dongonaeatu
dongonaeatu
OP
Original post by TGH1
No, as you haven't applied the log rules correctly.

alnb = ln(b^a)
lnc - lnd = ln(c/d)


is the answer log 62.5
Reply 4
Avatar for raheem94
raheem94
13
Original post by TGH1
No, as you haven't applied the log rules correctly.

alnb = ln(b^a)
lnc - lnd = ln(c/d)


The OP looks to have been stuck on a C2 question so it was better not to use the natural log.
Reply 5
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Remember, alogb=logba and logb−loga=log(ba) \displaystyle alogb=logb^a \ and \ logb-loga=log\left(\frac{b}{a}\right)


is the answer log 62.5
Reply 6
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
is the answer log 62.5


Yeah :smile:
Reply 7
Avatar for The Mr Z
The Mr Z
13
Original post by dongonaeatu
is the answer log 62.5


No, 5log3 = log(3^5) = log(243)

you have done log(5^3) which is incorrect.
Reply 8
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Yeah :smile:


sorry to keep bothering you.
if log10 a=4.1 find the value of a (1mark)

is it just a=4.1/log10?
Reply 9
Avatar for dongonaeatu
dongonaeatu
OP
Original post by The Mr Z
No, 5log3 = log(3^5) = log(243)

you have done log(5^3) which is incorrect.


the question was 3log5 not 5log3
Reply 10
Avatar for The Mr Z
The Mr Z
13
Original post by dongonaeatu
the question was 3log5 not 5log3


good point, in which case also check the second line of your first post - it's good to be consistent

then log 62.5 is correct
Reply 11
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
sorry to keep bothering you.
if log10 a=4.1 find the value of a (1mark)

is it just a=4.1/log10?


Remember, logab=x  ⟹  ax=b \displaystyle log_ab=x \implies a^{x} = b
Reply 12
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Remember, logab=x  ⟹  ax=b \displaystyle log_ab=x \implies a^{x} = b


10^4.1=a?

or is it log10^4.1=a
Reply 13
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
10^4.1=a?


Correct.
Reply 14
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
Correct.


if i do 10^4.1= 12589.25412

if i do log10^4.1= 0.6127838567 which one is it
Reply 15
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
if i do 10^4.1= 12589.25412

if i do log10^4.1= 0.6127838567 which one is it


log10a=4.1  ⟹  104.1=a=12589.25412 \displaystyle log_{10}a=4.1 \implies 10^{4.1} = a = 12589.25412
Reply 16
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
log10a=4.1  ⟹  104.1=a=12589.25412 \displaystyle log_{10}a=4.1 \implies 10^{4.1} = a = 12589.25412


thanks
Reply 17
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
thanks


No problem :smile:
Reply 18
Avatar for dongonaeatu
dongonaeatu
OP
Original post by raheem94
No problem :smile:


evaluate log3 1/9.... im pretty sure this is -2? as 3^2 is 9 and because its 1/9 its am minus so am i correct in saying the answer is -2
Reply 19
Avatar for raheem94
raheem94
13
Original post by dongonaeatu
evaluate log3 1/9.... im pretty sure this is -2? as 3^2 is 9 and because its 1/9 its am minus so am i correct in saying the answer is -2


log3(19)=log33−2=−2log33=−2×1=−2 \displaystyle log_3\left(\frac19\right) = log_33^{-2} = -2log_33 = -2 \times 1=-2

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