How do you solve #6^(x-2)=4^x#?

2 Answers
Nov 8, 2016

Answer:

#x=8.838#

Explanation:

In the equation #6^(x-2)=4^x#, taking log on both sides we get

#(x-2)log6=xlog4#

or #xlog6-2log6=xlog4#

or #xlog6-xlog4=2log6#

or #x=(2log6)/(log6-log4)#

= #(2xx0.7782)/(0.7782-0.6021)#

= #1.5564/0.1761=8.838#

Nov 8, 2016

Answer:

#x=8.838#

Explanation:

#6^(x-2)=4^x#

Log both sides.

#log 6^(x-2)=log 4^x#

Recall the log rule #logx^a = alogx#

#(x-2) log6 = xlog4#

Distribute the #log6#

#xlog6-2log6=xlog4#

Subtract #xlog6# from both sides.

#-2log6 = xlog4 - xlog6#

Factor out the #x# on the right side

-2log6=x(log4-log6)

Recall the log rule #loga - logb= log (a/b)#

#-2log6=xlog(4/6)#

#-2log6=xlog(2/3)#

Divide both sides by #log (2/3)#

#frac{-2log6}{log (2/3)}=x#

Use a calculator....

#x=8.838#