help with a logarithm series.

[roloddt]

We were given these logarithm series:
(I'll put the log bases in parentheses)

1) log(4) 64, log(8) 64, log(32) 64
2) log(7) 49, log(49) 49, log(343) 49
3) log(1/5) 125, log(1/125) 125, log(1/625) 125
4) log(8) 512, log(2) 512, log(16) 512

The answers are:
1) 3, 2, 6/5
2) 2, 1, 3/2
3) -3, -1, -3/4
4) 3, 9, 9/4

we're to figure out a trend to figure out the answer for the third term of each series judging from the answers of the first 2 terms.
I don't know how to figure that out and if you would, please show how you can get the third answer~
thank you in advance~

 

[galactus]

Here is the first one using the change of base formula.

The change of base formula is \(\displaystyle log_{a}(b)=\frac{log(b)}{log(a)}\)

\(\displaystyle log_{4}(64)=\frac{log(64)}{log(4)}=\frac{log(2^{6})}{log(2^{2})}=\frac{6\rlap{log(2)}////}{2\rlap{log(2)}////}=3\)

\(\displaystyle log_{8}(64)=\frac{log(64)}{log(8)}=\frac{log(2^{6})}{log(2^{3})}=\frac{6\rlap{log(2)}////}{3\rlap{log(2)}////}=2\)

\(\displaystyle log_{32}(64)=\frac{log(64)}{log(32)}=\frac{log(2^{6})}{log(2^{5})}=\frac{6\rlap{log(2)}////}{5\rlap{log(2)}////}=\frac{6}{5}\)


Now, see the patterns and how to proceed now?.

 

[roloddt]

oh ok~ thank you very much~
that saved me!

 

[roloddt]

Here is the first one using the change of base formula.

The change of base formula is \(\displaystyle log_{a}(b)=\frac{log(b)}{log(a)}\)

\(\displaystyle log_{4}(64)=\frac{log(64)}{log(4)}=\frac{log(2^{6})}{log(2^{2})}=\frac{6\rlap{log(2)}////}{2\rlap{log(2)}////}=3\)

\(\displaystyle log_{8}(64)=\frac{log(64)}{log(8)}=\frac{log(2^{6})}{log(2^{3})}=\frac{6\rlap{log(2)}////}{3\rlap{log(2)}////}=2\)

\(\displaystyle log_{32}(64)=\frac{log(64)}{log(32)}=\frac{log(2^{6})}{log(2^{5})}=\frac{6\rlap{log(2)}////}{5\rlap{log(2)}////}=\frac{6}{5}\)


Now, see the patterns and how to proceed now?.

I read the question wrong. It asks me to find the answer for the third term USING the answers from the first term and the second term.

 

[galactus]

Look at the pattern. The third term with respect to the first two terms is simply \(\displaystyle \frac{ab}{a+b}\)

a is the first term and b is the second.

BTW, for that second one, the third answer should be 2/3, not 3/2

 

[mmm4444bot]

I read the question wrong. It asks me to find the answer for the third term USING the answers from the first term and the second term.

If you are actually supposed to move from the first two terms (along with their given values) to the third without using "shortcuts", such as patterns, then I think the following might do.

First, note that the product of the bases in the first two terms equals the base in the third term.

(4)(8) = 32

(7)(49) = 343

(1/5)(1/125) = 1/625

(8)(2) = 16

This clues us in to looking for some way that involves the product.

I'll use the first exercise as an example.

Begin by switching to exponential notation for the first two terms with their given values.

log_4(64) = 3 becomes 4^3 = 64

log_8(64) = 2 becomes 8^2 = 64

Solve each of these exponential equations for the base (i.e., raise both sides to the reciprocal of the exponent).

4 = 64^(1/3)

8 = 64^(1/2)

Now comes the multiplication.

(4)(8) = 64^(1/3) * 64^(1/2)

32 = 64^(1/3 + 1/2)

32 = 64^(5/6)

Since the third term has 32 as the base, we rewrite with an exponent on 32 instead. (Again, raise both sides to the reciprocal of the exponent.)

32^(6/5) = 64

Switching back to logarithmic form shows the desired term because log_32(64) = 6/5.

(PS: There are no "series" in these exercises.)