Question

1)7 2log49 4. 2)log3 log2 8

Answer 1

1) 2 log 49 = 4             (49 = 7²)
    осн-е 7
2)   log   log 8  =  log 3     =1   (8 = 2³)
  осн-е3  осн-е2  осн-е 3

Answer 2

We can use the logarithmic identity log a^b = b*log a to simplify this expression:

7 * 2log49 4 = 7 * 2 * log 4 / log 49

Since 49 is 7^2, we can simplify log 49 as 2*log 7:

7 * 2log49 4 = 7 * 2 * log 4 / (2*log 7)

Simplifying further, we can cancel out the 2s:

7 * 2log49 4 = 7 * log 4 / log 7

So the final simplified form is:

7 * 2log49 4 = 7log 4 - 7log 7

log3 log2 8 can be rewritten using the change of base formula:

log3 log2 8 = log 8 / log 2 / log 3

Using the fact that log 8 = log 2^3 = 3*log 2:

log3 log2 8 = 3*log 2 / log 2 / log 3

Simplifying further by canceling out the log 2 terms:

log3 log2 8 = 3 / log 3