Solve for x.
ln(x+3) = ln10 - ln(x)

x =2

Step-by-step explanation:

Solve for x:

log(x + 3) = log(10) - log(x)

Hint: | Move everything to the left hand side.

Subtract log(10) - log(x) from both sides:

-log(10) + log(x) + log(x + 3) = 0

Hint: | Combine logarithms.

-log(10) + log(x) + log(x + 3) = log(1/10) + log(x) + log(x + 3) = log(1/10 x (x + 3)):

log(1/10 x (x + 3)) = 0

Hint: | Eliminate the logarithm from the left hand side.

Cancel logarithms by taking exp of both sides:

1/10 x (x + 3) = 1

Hint: | Multiply both sides by a constant to simplify the equation.

Multiply both sides by 10:

x (x + 3) = 10

Hint: | Write the quadratic polynomial on the left hand side in standard form.

Expand out terms of the left hand side:

x^2 + 3 x = 10

Hint: | Take one half of the coefficient of x and square it, then add it to both sides.

Add 9/4 to both sides:

x^2 + 3 x + 9/4 = 49/4

Hint: | Factor the left hand side.

Write the left hand side as a square:

(x + 3/2)^2 = 49/4

Hint: | Eliminate the exponent on the left hand side.

Take the square root of both sides:

x + 3/2 = 7/2 or x + 3/2 = -7/2

Hint: | Look at the first equation: Solve for x.

Subtract 3/2 from both sides:

x = 2 or x + 3/2 = -7/2

Hint: | Look at the second equation: Solve for x.

Subtract 3/2 from both sides:

x = 2 or x = -5

Hint: | Now test that these solutions are correct by substituting into the original equation.

Check the solution x = -5.

log(x + 3) ⇒ log(3 - 5) = i π + log(2) ≈ 0.693147 + 3.14159 i

log(10) - log(x) ⇒ log(10) - log(-5) = -i π + log(2) ≈ 0.693147 - 3.14159 i:

So this solution is incorrect

Hint: | Check the solution x = 2.

log(x + 3) ⇒ log(3 + 2) = log(5) ≈ 1.60944

log(10) - log(x) ⇒ log(10) - log(2) = log(5) ≈ 1.60944:

So this solution is correct

Hint: | Gather any correct solutions.

The solution is:

x = 2

idk

step-by-step explanation:

idk

find the magnitude of the trig term  

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(

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−

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by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

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find the upper bound of the range.

tap for more

y

=

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the range is  

y

≤

−

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or  

y

≥

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.

(

−

∞

,

−

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]

∪

[

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∞

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{

y

|

y

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step-by-step explanation: