log_2(x+1)=log_4(x^2-x+4)log2(x+1)=log4(x2−x+4) What is the value of x?

1 Answer
Cesareo R.
Aug 3, 2017

x=1x=1

Explanation:

Equivalently we have

(log_2(x+1))/(log_4 (x^2-x+4)) =( (log_e(x+1)/log_e 2))/( (log_e(x^2-x+4)/log_e4)) = 2(log_e(x+1)/ (log_e(x^2-x+4)))=1log2(x+1)log4(x2−x+4)=(loge(x+1)loge2)(loge(x2−x+4)loge4)=2(loge(x+1)loge(x2−x+4))=1

or

2log_e(x+1)=log_e(x^2-x+4)2loge(x+1)=loge(x2−x+4) or

(x+1)^2=(x^2-x+4)(x+1)2=(x2−x+4) or

2x+1=-x+42x+1=−x+4 or

3x=33x=3 or x = 1x=1

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