Solve algebraically, showing each step of your working, the equation #(8^(x-1) )^2-18(8^(x-1) )+32=0#??

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Answer 1 · 2018-02-24T21:05:36

#x=4/3# or #x=7/3#

What we have here is a quadratic equation in terms of #8^(x-1)#

Let #u=8^(x-1)#

The equation becomes

#u^2-18u+32=0#

#(u-16)(u-2)=0#

#u=16# or #u=2#

#8^(x-1)=2# or #8^(x-1)=16#

Considering #8^(x-1)=2# first;

In log form;

#log_8(2)=x-1#
#x-1=1/3#
#x=4/3#

Consider now #8^(x-1)=16#

In log form;

#log_8(16)=x-1#
#x-1=4/3#
#x=7/3#

So #x=4/3# or #x=7/3#