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i feel really stupid...log help?
so i know i should really know how to do this but for the life of me i can't remember. can someone please help me......i need to know the answer and show all my work. thanks!
find the sum: Log2 8 + log4 8 + log8 8 + log16 8 + log32 8
(the 2, 4, 8, 16, and 32 are supposed to be written slightly below the log....like the opposite way you would write an exponent.)
4 Answers
- jenh42002Lv 71 decade agoFavorite Answer
log2 8 = 3, because 2^3 = 8
log 4 8 = x, then 4^x = 8. 4 is 2^2, so 4^x = 2^2x = 8
2^2x = 2^3
2x = 3
x = 1.5
log8 8 = 1, since 8^1 = 8
log16 8 = x
16^x = 8
2^4x = 2^3
4x = 3
x = 3/4
log32 8 = x
32^x = 8
2^5x = 8
5x = 8
x = 8/5
So we have 3 + 1.5 + 1 + 3/4 + 8/5 = 7.85
- 1 decade ago
convert each log to an exponent...like 2 to the x equals 8....x would equal 3 (for the first log).
For log4 8..its 4 to the x equals 8, but make 4 = 2 raised to the second power and 8 = 2 raised to the third power...so it would be 2 raised to the 2x power equals 2 raised to the third power. then set the exponents equal to each other to get 3/2. Solve the other logs the same way
Then add up all the answers...should be 6.85.
- alphaLv 71 decade ago
The logarithm of any real number with respect to a given base is the index of the power to which the base is to be raised to be equal to the given number.
The given exp.=3+3/2+1+2/3+4/5=6 19/30
- 1 decade ago
maybe you should search websites for help because i've already forgotten Log. and i bet almost everyone else has too...
