Help! Weird Logarithm Problem!?

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Darkmoon

New member
It says "calculate the following" and give answers in the form, p/q where p and q are integers...

This is, i HAVE no idea what they really mean by 'calculate'... seriously. on my previous questions when they asked this i went on to find the expression without really knowing... Is that right?

1) log[base 1/5] 125, log [base 1/125] 125, log [base 1/625] 125
2) log[base 8] 512, log[base 2] 512, log[base 16] 512

Ok, i would have found the answer if the pattern wasn't so weird. The base thing doesn't fully match, but i give you my word. ThAT is how it looked like on the problem.

It then went on to say
"Describe how to obtain the 3rd answer in each row from the first 2 answers."

My teacher told me that the 3rd answer is really the 4th and the first 2 is really the first 3... which obviously made me more confused. Help?
 
the pattern really isn't so weird, u know...

1) log(base 1/5) 125 is same as [log 125]/[log 1/5] in any base!! and this is same as
[3log 5]/[- log 5] = -3 .... . .. .. cuz 125 = 5^3 and 1/5 = 5^(-1)
similarly, the next one wud be -1 and the one after that wud be -3/4

to get the third answer frm first two, here's wat u do... a is first answer, b is second and c is third:::
1/a + 1/b = 1/c ===> c = ab/(a+b)

2) this the same as above.. they just changed the values...
 
Here log_a means log(basea), ok??

Use this;
log_a b = [log_c b] / [log_c a]
and
log_a b^c = clog_a b

so
1.)
log_(1/5) 125
= [log_5 125] / [log_5 (1/5)]
= [log_5 5^3] / log_5 5^-1]
=3[log_5 ] / -1log_5 5
= 3/-1
=-3

log_(1/125) 1125
= log_5 (125) / log_5 (1/125)
= [log_5 5^3] / [log_5 5^(-3)]
= 3[log_5 5] / -3[log_5 5]
= 3/-3
= -1

log_(1/625) 125
= [log_5 125] / [log_5 (1/625)]
= [log_5 5^3] / [log_5 5^(-4)]
= 3/-4
= -(3/4)
============================================
Or u can do them really easily
log_(1/5) 125
125 is the 3rd power of 5
125 = 5^3
125 = 1/5^(-3)
so
log_(1/5) 125 = -3

log_(1/125) 125
125 is the 1st power of 125
125 = 125^1
125 = 1/125^-1
so
log_(1/125) 125 = -1

log_(1/625) 125
125 is the 3rd power of 5
125 = 5^3
625 is the 4th power of 5
625 = 5^4
5 = 625 ^(1/4)
so 125 = 5^3 = [(625)^(1/4]^3
125 = 625^(3/4)
125 = 1/625^(-3/4)
log_(1/625) 125 = -3/4
==============================================
how to obtain 3rd answer from 1 and 2;

log_(1/625) 125 = -3/4
log_1/5 125 = -3
log_(1/125) 125 = -1
so
-3/4 = - [ log_(1/5) 125] / [4log_(1/125) 125]
==============================================
2.)
log_8 512
= [log_2 512] / [log_2 8]
= [log_2 2^9] / [log_2 2^3]
= 9/3
=3

log_2 512
= log_2 2^9
= 9log_2 2
= 9

log_16 512
= [log_2 512] / [log_2 16]
= [log_2 2^9] / [log_2 2^4]
= 9/4
================================================
or easily
log_8 512
8 = 2^3
512 = 2^9
512 = 8^3
so
log_8 512 = 3

log_2 512= 9

log_16 512
16 = 2^4
2 = 16^(1/4)

512=2^9
512 = [16^(1/4)^9
512 = 16^9/4
so log_16 512 = 9/4
 
1) log[base 1/5] 125 = log[base 1/5] (1/5)^(-3) = -3 = -3/1

log [base 1/125] 125 = log [base 1/125] 125^(-1) = -1 = -1/1

log [base 1/625] 125 = {log [base 5] 125 } / {log [base 5] 1/625 } = 3/(-4) = -3/4

2) log[base 8] 512 = 3, since 8^3=512
log[base 2] 512 = 9, since 2^9=512
log[base 16] 512 = 9/4, since 16^(9/4) = (2^4)^(9/4) = 2^9 =512
 
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