Solving equations with logs and lx.?

Question

I have a TI-83 Plus and would like to know (how to solve) these.

1. Solve the equation:

Logx + Log(2x + 1) = 1


2. Solve the equation lnx - lnx(x-3) = ln3




The answer for 1. is 2
The answer for 2. is 9/2 or 4.5

Answer 1

log x + log (2x + 1) = 1
log x(2x + 1) = 1
x(2x + 1) = 10^1 = 10
2x^2 + x - 10 = 0
(2x + 5)(x - 2) = 0

x = -5/2 (throw out, cuz you can't do log of a neg number) or
x = 2.

ln x - ln (x-3) = ln 3
ln x/(x-3) = ln 3
x/(x - 3) = 3
x = 3(x - 3) = 3x - 9
9 = 2x
x = 9/2 = 4.5

You mistyped the second problem....

Answer 2

Solving equations with logs and lx.?
I have a TI-83 Plus and would like to know (how to solve) these.

1. Solve the equation:

Logx + Log(2x + 1) = 1
log(x(2x+1))=1
2x^2+x=10
2x^2+x-10=0
(2x+5)(x-2)=0
2x+5=0
x=-2.5
x-2=0
x=2

answers are 2, -2.5

For x=-2.5 as written originally, you will have the logs of 2 negative numbers, -2.5 & -4 which oh course aren't real, but by combining them, you wave the log of a positive number, 10

2. Solve the equation lnx - lnx(x-3) = ln3
ln(x/(x-3))=ln 3
x/(x-3)=3
x=3(x-3)
x=3x-9
-2x=-9
x=9/2=4.5




The answer for 1. is 2
The answer for 2. is 9/2 or 4.5

Answer 3

To solve the first one, you need to know the log rule for multiplication.

This rule reveals:

log(a) + log(b) = log a TIMES b.

Logx = loga

Log(2x+1) = logb

Substitute the values for each log and use algebra to find the answer for problem 1.

I will only do question 1 for you.

x (2x + 1) = 1

2x^2 + x = 1

NOTE: 10^1 = 10

2x^2 + x - 10 = 0

Solve for x and you will get two answers:

-5/2 and 2.

We CANNOT take the log of a negative number. So, we reject the answer -5/2.

The only other answer is x = 2.

Guido

Answer 4

log(x) + log(2x+1) = 1

Your first step is to remember the log property that the sum of logs equals the log of the product of their insides. That is,
log(a) + log(b) = log(ab)

That's exactly what we do with the left and side.

log (x * (2x + 1) ) = 1

Now, we expand.

log (2x^2 + x) = 1

At this point, we convert to exponential form.
(1) the base of the log becomes the base of the exponent.
(2) the answer becomes the power
(3) what remains becomes the answer.

Let's do this in steps.
"the base of the log is the base of the exponent"

In this case, we're using log base 10.

10^? = ?

"the answer becomes the power"

10^1 = ?

"what remains becomes the answer"

10^1 = 2x^2 + x

So that's our equation we work with. Simplifying it a bit,

10 = 2x^2 + x
2x^2 + x - 10 = 0

(2x + 5)(x -2) = 0
Therefore,

2x + 5 = 0 --> 2x = -5 --> x = -5/2
x - 2 = 0 ---> x = 2

Therefore, x = -5/2, 2

HOWEVER, you can't assume both values will work because you can't take the log of a negative number. What you have to do is test both values for validity.

Test x = -5/2: log(-5/2) + log(-5 + 1) already yields a negative number, so we discard that result.

Test x = 2: This works, because we're not taking the log of any negative number.

Therefore the answer is x = 2.

Answer 5

1. log(x) + log(2x+1) = log[x(2x+1)] = 1
elevate each to the base
so that x(2x+1) = 10
2x^2 + x - 10 = 0
(2x+5)(x-2)=0
x= -5/2 or 2

2. ln(x)-ln[x(x-3)] = ln[x^2(x-3)] = ln(3)
elevate to e
x^2(x-3) = 3
x^3 -3x^2 - 3 = 0
I forgot how to solve the cubic, please forgive, but the answer is here.

Answer 6

there is two thanks to respond to the question: yet there is a million answer in it because that its an absolute question there is two aspects to the question useful and detrimental. a million.(5x - 8) = (x + 4) subtract the left area by x and upload an 8 to the right to save the x's in a unmarried position and the numbers in a unmarried position,you need to do vise versa in words of aspects. 2. u might want to get 4x = 12, divide both area by four to the answer for x it really is x = 3. opposite direction we you need to do cuz often times the answer isn't continually an identical a million. -(5x - 8) = -(x + 4) which might want to finally end up being -5x + 8 = -x - 4 2. do an identical cancellations and yo get -4x = -12, divide both area by -4 and also you nonetheless get x = 3 your done.

Answer 7

1)
Logx + Log(2x + 1) = 1
Log [x*(2x + 1)] = log 10
x*(2x + 1) = 10
2x² + x = 10
2x² + x - 10 = 0

x = [-1+-(1+4*2*10)^1/2] / (2*2)
x = [-1+-(81)^1/2] / 4
x = [-1+-9] / 4
x = 2
___________________________________________
2)
lnx - lnx(x-3) = ln 3
ln[x/(x-3)] = ln 3
x/(x-3) = 3
x = 3(x-3)
x = 3x-9
2x = 9
x = 4.5

₢

Answer 8

log(a) + log(b) = log (ab)
log(a) - log(b) = log(a/b)
hope that helps

oh and log(ab) = c is equivalent to 10^(c) = ab

so for number 1. log(x(2x+1)) =1
therefore 10=x(2x+1) at which point it is simple algebra and the quadratic formula

2. ln(x/(x(x+3)))=ln(3)
since ln is on both sides you can drop it and 1/(x+3)=3 and again you get simple algebra.