How do you solve #log _x 8 = -3#?

1 Answer
Özgür Özer
Dec 4, 2015

#x=1/2#

Explanation:

#log_x 8=-3#
#=>log_x 2^3=-3#
#3log_x 2=-3# (The logarithm of the #3^(rd)# power of a number is #3# times the logarithm of the number itself)
Divide both side into #1/3#, in order to simplify left side:
#1/3*3log_x 2=1/3*(-3)#
#=>log_x 2=-1#
#=>x^-1=2# (definition of logarithm)
Multiply both side by #x#:
#x*x^-1=x*2#
#=>2x=x^0# (If we take the product of two exponentials with the same nonzero base, we simply add the exponents)
Multiply both side by #1/2#, in order to simplify left side:
#1/2*2x=1/2*1#
#x=1/2#

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