How do you solve #2^(8x-3) = 5^(6x-9)# ?

1 Answer
Geoff K.
Feb 16, 2017

#x=(3-9log_2 5)/(8-6log_2 5)=(3log_5 2 -9)/(8log_5 2-6)#.

Explanation:

For #a>0#, if #a^b=a^c# then #b=c#.

#5=2^(log_2 5)#

Thus

#color(white)=>2^(8x-3)=5^(6x-9)#
#=>2^(8x-3)=(2^(log_2 5))^(6x-9)#
#=>2^(8x-3)=2^((log_2 5)(6x-9))#

#=>8x-3=(log_2 5)(6x-9)#
#=>8x-3=6xlog_2 5-9log_2 5#

#=>8x-6xlog_2 5=3-9log_2 5#
#=>x(8-6log_2 5)=3-9log_2 5#

#=>x=(3-9log_2 5)/(8-6log_2 5)#

Another way:

#2=5^(log_5 2)#

So

#color(white)=>2^(8x-3)=5^(6x-9)#

#=>5^((log_5 2)(8x-3))=5^(6x-9)#

#=>8xlog_5 2-3log_5 2 = 6x-9#

#=>x(8log_5 2-6)=3log_5 2 -9#

#=>x=(3log_5 2 -9)/(8log_5 2-6)#

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