Solving Logarithms equation help: log_7 2(x+7) - log_7 (x+1) = 0

[thebeanstalk]

Hello,

So I was stuck on doing these 2 logarithm equations.

* log₇2(x+7) - log₇(x+1) = 0
so what I did here is:
log₇2(x+7)/x+1 = 0

x+1 [ 7 = 2(x+7)/x+1 ] x+1 (what I did here is to multiply both sides by x+1)
and it becomes

7x+7 = 2x+14
5x = 7
x = 5/7 (But when I plug it on the calculator, it gives me an answer of 1 instead of 0)

Another equation is

* log₅(x+1)² = log₅(x²-3)
What I did here is, since the logs has the same base, I equate (x+1)² = (x²-3)

(x+1)^{2 =} x²+2x+1

x²+2x+1 = x²-3 (I move the right side on the left side to equal to 0)

2x + 4 = 0
2x = -4
x = -2

Please help me what I did wrong here please!
Thank you in advance

 

[ksdhart2]

While Denis' solution is more elegant, and I believe easier to follow, the method you tried for the first problem will work. You just made an error in applying the logarithm rules. The rule is as follows:

\(\displaystyle log_n\left(a\right)-log_n\left(b\right)=log_n\left(\dfrac{a}{b}\right)\)

Thus, applied to the problem, you'd get:

\(\displaystyle log_7\left(2\left[x+7\right]\right)-log_7\left(x+1\right)=log_7\left(\dfrac{2\left[x+7\right]}{x+1}\right)\)

Be sure you understand why this is different from what you did:

\(\displaystyle log_7\left(2\left[x+7\right]\right)-log_7\left(x+1\right)\ne \dfrac{log_7\left(2\left[x+7\right]\right)}{x+1}\)

 

[thebeanstalk]

Same as a^x = a^y : so x = y

So 2(x + 7) = x + 1

I applied this, but I get x = -13

2(x + 7) = x + 1
2x + 14 = x + 1
2x - x = 1 - 14

x = -13 (and it's cannot be because it's impossible to get logarithm of negative numbers.)

 

[Harry_the_cat]

Hello,

So I was stuck on doing these 2 logarithm equations.

* log₇2(x+7) - log₇(x+1) = 0
so what I did here is:
log₇2(x+7)/x+1 = 0

This should be:
\(\displaystyle log_7\frac{2(x+7)}{x+1}=0\)

\(\displaystyle \frac{2(x+7)}{x+1} = 7^0\)

\(\displaystyle \frac{2x+14}{x+1} = 1\)

\(\displaystyle 2x+14 = x+1\)

\(\displaystyle x=-13\)

BUT if \(\displaystyle x=-13\) then original expression in not defined (can't take the log of a negative number).

Therefore there is no real solution.

x+1 [ 7 = 2(x+7)/x+1 ] x+1 (what I did here is to multiply both sides by x+1)
and it becomes

7x+7 = 2x+14
5x = 7
x = 5/7 (But when I plug it on the calculator, it gives me an answer of 1 instead of 0)

Another equation is

* log₅(x+1)² = log₅(x²-3)
What I did here is, since the logs has the same base, I equate (x+1)² = (x²-3)

(x+1)^{2 =} x²+2x+1

x²+2x+1 = x²-3 (I move the right side on the left side to equal to 0)

2x + 4 = 0
2x = -4
x = -2

Please help me what I did wrong here please!
Thank you in advance

See red comment for first one.

Second one seems correct to me - although you should check that x=-2 substitutes back into the original equation without getting log of a non-positive number.
In this case it is fine.