Log2(x+1)=log4(x^2-x+4) What is the value of x?

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Answer 1 · 2018-04-18T04:00:06

#x=1#

#log_2(x+1)=log_4(x^2-x+4)#

Lets start by converting the #log_4# term to a #log_2# term. We can do this by using the change-of-base formula.

#log_ax=(log_bx)/(log_ba)#

So we can rewrite our expression as

#log_2(x+1)=(log_2(x^2-x+4))/(log_2(4))#

But #log_2(4)=2# so,

#log_2(x+1)=(log_2(x^2-x+4))/2#

Multiply both sides of this equation by 2.

#2log_2(x+1)=log_2(x^2-x+4)#

Now we can use the property of logarithms that says #alogx=logx^a#. This lets us rewrite our equation as

#log_2(x+1)^2=log_2(x^2-x+4)#.

If the logs are equal, then their arguments must be equal.

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

Expand the squared binomial on the left-hand side of this equation.

#x^2+2x+1=x^2-x+4#

Combine like terms.

#3x=3#

Divide both sides by 3.

#x=1#

Check to make sure this makes sense by plugging this value for #x# into the original equation.

#log_2(1+1)# =?= #log_4(1^2-1+4)#

#log_2(2)# =?= #log_4(4)#

#1=1#

Yup.