This one has lost me
2007-07-25 12:40:35 · 3 answers · asked by abercrombi_4_me 1 in Science & Mathematics ➔ Mathematics
yeah Using properties of logarithms, rewrite the ecpression below as a single logarithm!
2007-07-25 13:48:10 · update #1
7log [base 4] x+ log [base 4] (B/E) - log [base 4] a^2
=log_4[ (x^7*B/E)/a^2]
=log_4[ Bx^7/(Ea^2)]
2007-07-25 12:50:15 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
pardon me, but i ain't really know what your qn is about. do you mean simplify it? if it is, den here it is.
7log [base 4] x+ log [base 4] (B/E) - log [base 4] a^2
= log [base 4] x^7 + log [base 4] (B/E) - log [base 4]a^2
= log [base 4] [(Bx^7) / (Ea^2)]
2007-07-25 12:54:13 · answer #2 · answered by pearlysim 1 · 0⤊ 0⤋
i assume you meant to simplify
7 log_4 (x) + log_4 (B/E) - log_4 (a^2)
exponent rule: log a^x = x log a
log_4 (x^7) + log_4 (B/E) - log_4 (a^2)
product rule: log a + log b = log (ab)
log_4 (x^7 * B/E) - log_4 (a^2)
log_4 (Bx^7 / E) - log_4 (a^2)
quotient rule: log a - log b = log (a/b)
log_4 ( (Bx^7 / E) / a^2)
log_4 (Bx^7 / Ea^2)
2007-07-25 12:51:01 · answer #3 · answered by 7 · 0⤊ 0⤋