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# Please can someone help me explain this Log base 10 (X^²+4)=2 + log base 10 X - log base 10 20?

When solving, this was how the teacher started

- Log base 10 (X^²+4)+ log base 10 20 - log base 10 X = 2

- Log base 10 (X^²+4/X)20 = 2

Please am confuse, please can anyone help me out??

### 3 Answers

- MyRankLv 63 weeks ago
log₁₀(x²+4) = 2 + log₁₀x - log₁₀²⁰

-log₁₀(x²+4) + 2 + log₁₀x - log₁₀²⁰

-log₁₀(x²+4) + 2 + log₁₀(x/20) = 0

(∵ loga/b = loga-logb)

2+log₁₀((x/20) / x²+4)) = 0

log₁₀(x/20(x²+4)) = -2

x/(20(x²+4) = 10⁻²

x/20(x²+4) = 1/100

100x = 20(x²+4)

100x = 20x² + 80

20x² - 100x + 80 = 0

x²-5x+4 = 0

x²-4x-x+4 = 0

x(x-4)-1(x-4) = 0

(x-4)(x-1) = 0

x = 4, 1

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- mizooLv 73 weeks ago
Remember log_b a = c => a = b^c? You need to use it.

But there is an error in your solution:

log_10 20 is not equal to 2.

log_10 (x^2+4) = 2 + log_10 x - log_10 20

log_10 (x^2+4) = 2 + log_10 (x/20)

log_10 (x^2+4) - log_10 (x/20) = 2

log_10 (x^2+4)/(x/20) = 2

10^2 = (x^2+4)/(x/20)

20x^2 - 100x + 80 = 0

x^2 - 5x + 4 = 0

(x - 1)(x - 4) = 0

x = 1, 4

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- llafferLv 73 weeks ago
The general rules here are:

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

log(x) + log(y) = log(xy)

The difference of two logs of the same base is the same as the log of the quotient:

log(x) - log(z) = log(x/z)

In your case, you are adding and subtracting. I won't specify the base since "log" with no base shown implies base 10, so you have:

log(x² + 4) + log(20) - log(x) = 2

So we can multiply the first two and divide the third and put it all within a single log:

log[20(x² + 4) / x] = 2

Then to solve this you'd put both sides as an exponent over a base of 10 to cancel out the log, but I won't continue that here as I've answered the question.

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