How can this be solved??? help !!

#4^(x+1)-9*2^x+2!=0#

#4^(x+1)-9*2^x+2!=0#

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1s2s2p Share
Feb 9, 2018

Answer:

#x!=-2or1#

Explanation:

We have #4^(x+1)-9*2^x+2!=0#

#4^(x+1)=(2^(x+1))^2=2^(2x+2)#

Now we have:
#4(2^(2x))-9(2^x)+2!=0#

Let #2^x=y#

#4y^2-9y+2!=0#

Via the quadratic formula:
#y=(-b+-sqrt(b^2-4ac))/(2a)#

#y=(-(-9)+-sqrt((-9)^2-4(4*2)))/(2*4)#

#y=(9+-sqrt(81-32))/8#

#y=(9+-sqrt(49))/8#

#y=(9+-7)/8#

#y=1/4or2#

#2^x=1/4or2#

#x=log_2(1/4)orlog_2(2)#

#x=-2or1#

So, for #4^(x+1)-9*2^x+2!=0#, #x!=-2or1#

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