How do you solve #log_8 (1) + log_9 (9) + log_5 (25) + 3x= 6#?

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Answer 1 · 2016-01-27T00:42:33

I found #x=1#

Here we can take advantage of the definition of log:
#log_ax=y -> x=a^y#
so that we get:
#0+1+2+3x=6#
#3x=3#
and
#x=1#

Remember that:
#8^0=1#
#9^1=9#
#5^2=25#

Answer 2 · 2016-01-27T01:39:40

#x= 1#

To solve this problem, we need to remember severals logarithmic properties.

#log_a a = 1 # , given #a# is any positive number, #a>0#
#log_a 1= 0 #
#log_a a^n = n #

We have

#log_8 (1) + log_9(9) + log5(25) + 3x = 6#

#0 + 1 + log_5(5^2) + 3x =6#
#0 + 1 + 2 + 3x = 6#
Combine like terms

#3 + 3x = 6#

#3x = 3#

#x = 1#