Question

1)log(x+2)-log^4=log3x
2)log(x-1)+log^3=logx​

Answer 1

Answer:

Given

2logx=log2+log(3x−4)

Step 1: Rewrite the equation as a single logarithm on the right hand side, using the sum to product rule, like this

logx2=log(2⋅(3x−4))logx2=log(6x−8)

Step 2: Transform into the exponential form with the base of 10 like this (or most simple way to put it, if there are log with same base on each side of equation then we can "drop" the log).

10logx2=10log

(6x−8)x2=6x−8

Now, we can begin to solve the equation by subtracting

6x

and adding 8 to both sides to get

x2−6x+8=0

This can be solved by using a factoring method like this …

The factors of 8 that add up to -6 are -2 and -4.

(x−2)(x−4)=0

x−2=0

⇒x=2

or

x−4=0

⇒x=4

We will need to the check the solutions above to determine whether there are any extraneous solutions (solution that don't work (aka "check out").

Check

x=22log(2)=log2+log(3⋅2-4)2log2

=1log2+1log22log2=2log2

Therefore

x=2 is a solution.

Check x=42log(4)=log2+log(3⋅4−4)2log4

log2+log8log42=log(2⋅8)log16=log16

Therefore x=4

is also another solution.

Step-by-step explanation:

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Answer 2

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