What is the value of #x# if #log_6 48 = log_6(x + 7) + log_6(x - 1)#?

Question

1257 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-06T20:06:08

#{5}#

Combine the logarithms.

#log_6 48 = log_6 ((x + 7)(x - 1))#

If #log_an = log_am#, then #n = m#.

#48 = (x + 7)(x - 1)#

#48 = x^2 + 7x - x - 7#

#48 = x^2 + 6x- 7#

#0 = x^2 + 6x - 55#

#0 = (x+ 11)(x -5)#

#x = -11 and 5#

#x = -11# is clearly extraneous as a logarithm can never be negative. #x= 5# is valid, though.

Practice Exercises

Solve the following equations using #log_a n - log_a m = log_a(n/m)# and #log_a m + log_a n = log_a (m * n)#.

a) #log_2 (2x) + log_2 (1/2x) = log_2 64#
b) #log_3 (x + 2) - log_3 (x - 4) = 1#

Answers:

a) #{8}#
b) #{7}#

Hopefully this helps, and good luck!