could someone tell me how to solve log2 24- log2 3= log5 x?
2006-10-27 15:17:12 · 6 answers · asked by Nancy 1 in Science & Mathematics ➔ Mathematics
I will put the base number in square brackets for easier reading.
log[2]24 - log[2]3 = log[5]x
Therefore, log[2](24 / 3) = log[5]x, because logA - logB = log(A / B) (if to the same base)
That is, log[2]8 = log[5]x, because 24 / 3 = 8.
But log[2]8 = 3, because 2^3 = 8, by the definition of a logarithm, which states that if a^p = N, then p = log[a]N.
So, 3 = log[5]x
Therefore, x = 5^3, also by the definition of a logarithm.
Thus, x = 125.
2006-10-27 16:39:09 · answer #1 · answered by falzoon 7 · 0⤊ 0⤋
Fact 1: (log base a of b) - (log base a of c) = (log base a of b/c)
So log2(24) - log2(3) = log2(24/3) = log2(8) = 3
So log5(x) = 3, x = 5^3 = 125.
(I guess you don't need Fact 2 for this)
2006-10-27 15:33:23 · answer #2 · answered by sofarsogood 5 · 1⤊ 0⤋
Try to bring the left hand side to one log expression using the log rules, since it is a difference it follows logaB-logaC=loga(B/C)
As such log2 24- log2 3=log2 8=log8/log2=3
3=log5 x and such 5^3=x and x=125
plug back into orginal equation to verify: remeber another important log rule logaB=log10 B/log 10 a thereby allowing for calculator use
log 24/log 2 - log 3/log 2=(use calculator) 4.585-1.585=3
Now check right hand side:
log5 x(plug 125 for x as found)
log5 125=log125/log5=3
Ha,Ha both left and right dies match and as such x=125 is the true value of x
2006-10-27 20:23:33 · answer #3 · answered by Zidane 3 · 0⤊ 0⤋
log2 24-log2 3=log2 21
2006-10-27 16:08:02 · answer #4 · answered by peterwan1982 2 · 0⤊ 0⤋
log a - log b = log (a/b)
log2 24 - log2 3 = log5 x
log2 (24/3) = log5 x
log2 8 = log5 x
3 = log5 x
5^3 = x
x = 125
2006-10-27 16:15:10 · answer #5 · answered by M. Abuhelwa 5 · 0⤊ 0⤋
log2(24)-log2(3)=log2(24/3)=log2(8)=3
3=log5(x), that is x=5^3=125
2006-10-27 15:37:00 · answer #6 · answered by Anonymous · 0⤊ 0⤋