Solve for #x# when #log(x^2-x-6)+x=log(x+2)+4#?

Answer 1

#x=4#

#log(x^2-x-6)=log((x-3)(x+2))# #=log(x-3)+log(x+2)#
Therefore, #log(x-3)+log(x+2)+x=log(x+2)+4# #log(x-3)=4-x# #x-3=10^(4-x)#
Solutions to problems such as these cannot be calculated via regular methods, however, just looking at this equation, the answer can be seen to be #4#, as #4-3# is #1#, and #10^0# is also #1#.
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Answer 2

#x=4#

#log(x^2-x-6)+x=log(x+2)+4#

Move all the #log# functions to one side and the constants the other,

#log(x^2-x-6)-log(x+2)=4-x#

Make use of the laws of logarithms,

#log((x^2-x-6)/(x+2))=4-x#
#color(white)(xxxxx)log(x-3)=4-x#

Since #log(x-3)# and #4-x# are equal, plot them against #y#,

#color(purple)(y=log(x-3)#
#y=4-x#

Hence, #x=4#.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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