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log help needed

log10x^5+ 3log10 x^4

how do i do it whats the rule in the solutions its

logx^5 + log(x^4)^3

how did it get to this
Reply 1
or it it just the opposite?

for example logx^3
can be written 3logx

and i have equivelent to 3logx

so i can change that to logx^3??
Reply 2
Okay well log(a)^x=xlog(a)
Can you apply this to the question?
Reply 3
Original post by dongonaeatu
or it it just the opposite?

for example logx^3
can be written 3logx

and i have equivelent to 3logx

so i can change that to logx^3??

Yes you are right
Reply 4
Original post by dongonaeatu
log10x^5+ 3log10 x^4

how do i do it whats the rule in the solutions its

logx^5 + log(x^4)^3

how did it get to this


Just watch this video.
Reply 5
Original post by ad99797
Yes you are right


yes thank you. so it is logx17 final answer

only tricky bit is when you move the 3 from 3logx^4 where abouts to put the 3

but i guess its just log(x^4)^3 so thats logx^12
Reply 6
A*log(x) = log(x)^A
(edited 11 years ago)
Reply 7
Original post by raheem94
Just watch this video.


ok but is my answer logx17 right
Reply 8
Original post by dongonaeatu
ok but is my answer logx17 right


Yes, the answer is log10x17 log_{10} x^{17} which can also be written as 17log10x 17log_{10}x

:congrats:
Reply 9
Original post by raheem94
Yes, the answer is log10x17 log_{10} x^{17} which can also be written as 17log10x 17log_{10}x

:congrats:


the next question part ii is loga 1 -loga a^b

so loga 1- bloga a= the bases are the same so i can just divide them?

so its loga1/bloga a
Reply 10
Original post by raheem94
Yes, the answer is log10x17 log_{10} x^{17} which can also be written as 17log10x 17log_{10}x

:congrats:


actually i think i've done that wrong
Reply 11
Original post by raheem94
Yes, the answer is log10x17 log_{10} x^{17} which can also be written as 17log10x 17log_{10}x

:congrats:


loga^1=0 according to that video you sent me

so 0-loga a^b
Sorry to intrude. But when logs are added, aren't they instead multiplied? So shouldn't it be 12x5=60 log x.

Or am I missing something?
Reply 13
Original post by dongonaeatu
loga^1=0 according to that video you sent me

so 0-loga a^b


loga1logaab log_a1 - log_aa^b

You are correct that loga1=0 log_a1 =0

So we get, 0logaab=logaab=blogaa 0 - log_aa^b = -log_aa^b = -blog_aa

There is a rule, that logcc=1 log_cc = 1 , which means if the base of the log is same as the number in front of it, then it equals 1.

Hence, blogaa=b(1)=b -blog_aa = -b(1) = -b

Does this makes sense?
Reply 14
Original post by t00 g00d for u
Sorry to intrude. But when logs are added, aren't they instead multiplied? So shouldn't it be 12x5=60 log x.

Or am I missing something?


logx5+3logx4=logx5+log(x4)3=logx5+logx12=log(x5×x12)=logx5+12=logx17=17logx logx^5 + 3logx^4 = logx^5 + log(x^4)^3 = logx^5 + logx^{12} = log(x^5 \times x^{12}) = logx^{5+12} = logx^{17} = 17logx

Hope it makes sense.
Reply 15
Original post by raheem94
Yes, the answer is log10x17 log_{10} x^{17} which can also be written as 17log10x 17log_{10}x

:congrats:


well it is logx^5 + logx^12

and when multiplying indices which have the same base (e.g x) we add them

if we ignore the logs, its just like x^5+x^12. the rules of indices says we add these together.

Likewise if it was x^5 divided by x^12, we would do x^5-x^12= x^-7.
Original post by raheem94
logx5+3logx4=logx5+log(x4)3=logx5+logx12=log(x5×x12)=logx5+12=logx17=17logx logx^5 + 3logx^4 = logx^5 + log(x^4)^3 = logx^5 + logx^{12} = log(x^5 \times x^{12}) = logx^{5+12} = logx^{17} = 17logx

Hope it makes sense.


Ahh I see. Of course, thanks.
Original post by dongonaeatu
well it is logx^5 + logx^12

and when multiplying indices which have the same base (e.g x) we add them

if we ignore the logs, its just like x^5+x^12. the rules of indices says we add these together.

Likewise if it was x^5 divided by x^12, we would do x^5-x^12= x^-7.


Yeah I get the indice rule. I just thought that it was 5logx + 12 log x, so therefore timed those numbers.
Reply 18
Original post by raheem94
loga1logaab log_a1 - log_aa^b

You are correct that loga1=0 log_a1 =0

So we get, 0logaab=logaab=blogaa 0 - log_aa^b = -log_aa^b = -blog_aa

There is a rule, that logcc=1 log_cc = 1 , which means if the base of the log is same as the number in front of it, then it equals 1.

Hence, blogaa=b(1)=b -blog_aa = -b(1) = -b

Does this makes sense?


sorry, i understand up to -bloga a.
Reply 19
Original post by raheem94
logx5+3logx4=logx5+log(x4)3=logx5+logx12=log(x5×x12)=logx5+12=logx17=17logx logx^5 + 3logx^4 = logx^5 + log(x^4)^3 = logx^5 + logx^{12} = log(x^5 \times x^{12}) = logx^{5+12} = logx^{17} = 17logx

Hope it makes sense.


ohhh wait.

loga^1=0
loga^a=1

so its -btimes1

which is -b

thanks i got it

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