Apakah x jika log_3 (2x-1) = 2 + log_3 (x-4)?

Jawapan:

#x = 5 #

Penjelasan:

Kami akan menggunakan perkara berikut:

• #log_a (b) - log_a (c) = log_a (b / c) #
• # a ^ (log_a (b)) = b #

# log_3 (2x-1) = 2 + log_3 (x-4) #

# => log_3 (2x-1) - log_3 (x-4) = 2 #

# => log_3 ((2x-1) / (x-4)) = 2 #

# => 3 ^ (log_3 ((2x-1) / (x-4))) = 3 ^ 2 #

# => (2x-1) / (x-4) = 9 #

# => 2x - 1 = 9x - 36 #

# => -7x = -35 #

# => x = 5 #

Jawapan:

Saya jumpa: # x = 5 #

Penjelasan:

Kita boleh mula menulisnya sebagai:

# log_3 (2x-1) -log_3 (x-4) = 2 #

gunakan harta log: # logx-logy = log (x / y) # dan tulis:

# log_3 ((2x-1) / (x-4)) = 2 #

gunakan takrif log:

# log_bx = a-> x = b ^ a #

untuk mendapatkan:

# (2x-1) / (x-4) = 3 ^ 2 # menyusun semula:

# 2x-1 = 9 (x-4) #

# 2x-9x = -36 + 1 #

# 7x = 35 #

# x = 35/7 = 5 #