Log problem, solve for x

[asissa]

Here is the problem: log[sub:2n2r5crv]2[/sub:2n2r5crv](x-4) + log[sub:2n2r5crv]2[/sub:2n2r5crv]x = 5
There are two of us working the problem, and we have different answers. Would you please review and let us know which is correct. If neither are correct, can you show us how to correctly solve this problem? Thank you.

Answer 1
log[sub:2n2r5crv]2[/sub:2n2r5crv](x)(x-4)=5^2
log[sub:2n2r5crv]2[/sub:2n2r5crv](x^2 - 4x) = 5
2^5 = x^2-4x
32 = x^2 - 4x
x^2 - 4x - 32 = 0
(x + 4)(x - 8)
x = -4, 8

Answer 2
log[sub:2n2r5crv]2[/sub:2n2r5crv](x-4) + log[sub:2n2r5crv]2[/sub:2n2r5crv]x = 5
log[sub:2n2r5crv]2[/sub:2n2r5crv](x-4) = 5 + log[sub:2n2r5crv]2[/sub:2n2r5crv]x
log[sub:2n2r5crv]2[/sub:2n2r5crv](x-4) / log[sub:2n2r5crv]2[/sub:2n2r5crv]x = 5
x^2 - 4x = 5x
x^2 = 9x
x^2 - 9x = 0
x(x + 9) = 0
x = 0, 9

We have looked at this and are now just going around in circles. Help!

 

[tkhunny]

Until you learn to check your answer and to be careful, you will be going round and round.

log2(x-4) + log2(x) = 5

log rules

log2[(x-4)(x)] = 5

log-exponent equivalence

(x-4)(x) = 2^5 = 32

Quadratic Equation from here. Let's see what you get.

 

[mmm4444bot]

x^2 - 4x - 32 = 0
(x + 4)(x - 8)
x = -4, 8

This quadratic equation looks good, and the candidates for x look correct.

Here's something to remember: Logarithms of negative numbers are meaningless within the Real-number system.

You can verify the other candidate by substituting it for x in the given equation and confirming that both sides have the same numerical value.


Answer 2

log[sub:15mtaj9j]2[/sub:15mtaj9j](x-4) = 5 + log[sub:15mtaj9j]2[/sub:15mtaj9j]x

log[sub:15mtaj9j]2[/sub:15mtaj9j](x-4) / log[sub:15mtaj9j]2[/sub:15mtaj9j]x = 5

This step is not correct. If we divide both sides of an equation by something, each and every term in the equation (both sides) must be divided.

After dividing both sides of the first equation above by log[sub:15mtaj9j]2[/sub:15mtaj9j]x, the righthand side would not be 5:

log[sub:15mtaj9j]2[/sub:15mtaj9j](x-4)/log[sub:15mtaj9j]2[/sub:15mtaj9j]x = 5/log[sub:15mtaj9j]2[/sub:15mtaj9j]x + 1

Getting a single logarithm on one side and then switching to exponential form to obtain a quadratic equation -- as shown in the first approach -- is the way to go.

 

[innosods]

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Here is the problem: log2(x-4) + log2x = 5There are two of us working the problem, and we have different answers. Would you please review and let us know which is correct. If neither are correct, can you show us how to correctly solve this problem? Thank you.Answer 1log2(x)(x-4)=5^2log2(x^2 - 4x) = 52^5 = x^2-4x32 = x^2 - 4xx^2 - 4x - 32 = 0(x + 4)(x - 8)x = -4, 8Answer 2log2(x-4) + log2x = 5log2(x-4) = 5 + log2xlog2(x-4) / log2x = 5x^2 - 4x = 5xx^2 = 9xx^2 - 9x = 0x(x + 9) = 0x = 0, 9We have looked at this and are now just going around in circles. Help!

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