Science & Mathematics » Mathematics » Math help. Ln(4)+Ln (4×^2-4)=Ln (20) Solve for x.?

# Math help. Ln(4)+Ln (4×^2-4)=Ln (20) Solve for x.?

• x = +1.5 or -1.5
Ln (4) + Ln(4x^2 -4) = Ln (20)
Ln (16x^2 - 16) = Ln(20)
16x^2 - 16 = 20
x^2 =36/16 = 2.25
x = +1.5 or -1.5
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• ln(4) + ln(4x² - 4) = ln(20)

The sum of two logs of the same base is the same as the log of the product, so:

ln[(4(4x² - 4)] = ln(20)

We now have two values equal to each other with the log of the same base on both sides. So what's inside must also be the same, so:

4(4x² - 4) = 20

Divide both sides by 4, then add 4 to both sides:

4x² - 4 = 5
4x² = 9

Square root of both sides:

2x = ±3

x = ±3/2

testing both solutions to make sure that are in the domain of the equation:

ln(4) + ln(4x² - 4) = ln(20)
ln(4) + ln(4(-3/2)² - 4) = ln(20) and ln(4) + ln(4(3/2)² - 4) = ln(20)
ln(4) + ln(4(9/4) - 4) = ln(20) and ln(4) + ln(4(9/4) - 4) = ln(20)

Now that both equations are the same, I'll just solve for one of them:

ln(4) + ln(9 - 4) = ln(20)
ln(4) + ln(5) = ln(20)
1.38629 + 1.60944 = 2.99573
2.99573 = 2.99573
TRUE

So it does work out for both values for x

Again, the solution is:

x = ± 3/2

Hope this helped. If it did, please give best answer.
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