How do you solve #log_4(x) + log_4(x+2) = 1/2 log_4 9#?

3 Answers

#x=1# and #x=-3#

Explanation:

Start from the given equation:

#log_4 x + log_4 (x+2)= 1/2 log_4 9#

#log_4 x(x+2)= log_4 9^(1/2)#

#Antilog(log_4 x(x+2))= Antilog( log_4 9^(1/2))#

# x(x+2)= 9^(1/2)#

# x(x+2)= 3#

#x^2+2x-3=0#

#(x+3)(x-1)=0#

#x+3=0# and #x-1=0#

#x=-3# and #x=1#

Have a nice day from the Philiipines !!!

Feb 4, 2016

x = 1

Explanation:

Using the following 'laws of logs'

#• logx + logy = logxy#

#• logx^n = nlogx #

then # log_4 x + log_4(x+2) = log_4 9^(1/2) #

hence # x ( x+ 2 ) = 9^(1/2) = 3 #

( since if # log_b a = log_b c color(black)(" then ") a = c ) #

so x (x+ 2 ) = 3 → # x^2 + 2x - 3 = 0 #

Factors to give : (x + 3 )(x - 1 ) = 0

# rArr x = - 3 , x = 1 #

but x ≠ - 3 hence x = 1

#x=1# and #x=-3#

Explanation:

For #x=-3# is also a solution considering that
the absolute value of x and (x+2) in the equation may be used.

example:

#log_4 abs(x)+log_4 abs(x+2)=1/2*log_4 9#

#log_4 abs(-3)+log_4 abs(-3+2)=1/2*log_4 9#

#log_4 3+log_4 1=1/2*log_4 9#

#log_4 3+ 0=log_4 9^(1/2)#

#log_4 3=log_4 3#

therefore #x=-3# is a solution also !