How do you solve #log_6 x -log_6(x-6)=1#?

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Answer 1 · 2017-01-09T16:11:10

#x=7.2#

Using the Rule# : log_ma-log_mb=log_m(a/b)#, we get,

#log_6(x/(x-6))=1...............(1)#

Since, #log_ma=n rArr m^n=a#, we have, from #(1)#,

#x/(x-6)=6^1=6#

#:. x=6(x-6)=6x-36#

#:. 5x=36 rArr x=36/5=7.2#

This root satisfy the given eqn.

Hence, #x=7.2# is the Soln.

Answer 2 · 2017-01-09T16:18:11

Use #log(a) - log(b) = log(a/b)# to combine to a single log, eliminate the log by making both sides a power of the base, then solve for x.

Use the property #log(a) - log(b) = log(a/b)#:

#log_6(x/(x - 6)) = 1#

Make the log disappear by making both sides the exponent of 6:

#x/(x - 6) = 6^1#

Solve for x:

#x = 6x - 36#

#-5x = -36#

#x = 36/5#

Check:

#log_6(36/5) - log_6(36/5 - 6) = 1#

#log_6(36) - log_6(5) - log_6(6) + log_5(5) = 1#

#2log_6(6) - log_6(5) - log_6(6) + log_5(5) = 1#

#log_6(6) = 1#

#1 = 1#

This checks.