Solve the equation and check for extraneous solutions.

Question

Solve the equation. Check for extraneous solutions.

In(x^2+3)=In(4)

log_(2)(x^2-5)=2/3

Answer 1

if ln(x) = ln(4) does it make sense to you the x = 4?

We can then extrapolate that property to ln(x² + 3) = ln(4) to mean that x² + 3 = 4

x² + 3 = 4

x² = 1

x = ±1

Try x = 1 into ln(x² + 3) = ln(4)

ln((1)² + 3) = ln(4)

ln(4) = ln(4)

So x = 1 is a solution

Try x = -1 into ln(x² + 3) = ln(4)

ln((-1)² + 3) = ln(4)

ln(4) = ln(4)

So x = -1 is a solution

For the second problem, I'm assuming it is: log₂(x² - 5) = 2/3

To convert a log problem into an exponential problem we can remember that a logarithm is an exponent so log₂(x) = y ("y" is the exponent" and "2" is the base) means 2^y = x

Therefore extrapolating that property to your problem:

log₂(x² - 5) = 2/3 means 2^(2/3) = x² - 5

2^(2/3) = x² - 5

(2²)^1/3 = x² - 5

(4)^1/3 = x² - 5

x² = 5 + (4)^1/3

we can approximate 4^1/3 as 1.587401052 using a calculator

x² = 6.587401052

x = ± √6.587401052

x = ±2.566593277

Trying 2.566593277 into log₂(x² - 5) = 2/3:

log₂((2.566593277)² - 5) = 2/3

log₂(6.587401052 - 5) = 2/3

log₂(1.587401052) = 2/3

0.6666666667 = 2/3

So x ≈ 2.566593277 is a solution

Trying -2.566593277 into log₂(x² - 5) = 2/3:

log₂((-2.566593277)² - 5) = 2/3

log₂(6.587401052 - 5) = 2/3

log₂(1.587401052) = 2/3

0.6666666667 = 2/3

So x ≈ -2.566593277 is a solution