How do you solve #2^x = 3^(x-1) #?

1 Answer
Dec 11, 2015

#x=log_(2/3)3#

Explanation:

#color(white)(xxx)2^x=3^(x-1)#

#=>log_3 2^x=x-1#

The logarithm of the #x^(th)# power of a number is #x# times the logarithm of the number itself:
#color(white)(xxx)xlog_3 2=x-1#

Multiply both sides by #color(red)(1/x)#
#color(white)(xxx)color(red)(1/x*)xlog_3 2=color(red)(1/x*)(x-1)#
#=>(x-1)/x=log_3 2#

Add #color(red)(-1)# to both sides:
#color(white)(xxx)(x-1)/xcolor(red)(-1)=log_3 2color(red)(-1)#
#=>1/x=log_3 2-1#
#=>1/x=log_3 2-color(blue)(log_3 3)color(white)(xxxxx)# (because for #AAainRR, a^1=a#)

The logarithm of the ratio of two numbers is the difference of the logarithms:
#=>1/x=log_3 (2/3)#
#=>x=log_(2/3)3#