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The question asks:

Solve $$\log_5(x-1) + \log_5(x-2) - \log_5(x+6)= 0 $$

I know that according to log laws, addition with the same base is equal to multiplication and subtraction is equal to division (and vice versa)

By doing this I get $$\log_5{(x-1)(x-2)}=0 (x+6)$$

By moving the $x+6$ to the other side it should become zero, ($x+6 \times 0$), however the answer in the textbook moves the $x+6$ to the other side of the equation and then solves. Why am I wrong? And when would you do my method then?

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  • $\begingroup$ To answer your original question: the book is adding $\log_5(x+6)$ to both sides. You wouldn't multiply the equation $x - 2 = 0$ by $2$ to solve for $x$, would you? Same thing here. $\endgroup$ – pjs36 Mar 10 '16 at 19:25
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By the same rules you mentioned you should get $$\log_5 \left[(x-1)(x-2)\right]=\log_5[x+6]$$ which after exponentiation with $5$ becomes $$(x-1)(x-2)=x+6$$

If you want to do it by including all three terms in the logarithm, you get the same: $$\log_5 \left[\frac{(x-1)(x-2)}{x+6}\right]=0$$ but now you can't just multiply by $x+6$ on both sides and expect it to disappear in the denominator in the $\log$, since it is inside the $\log$. What you have to do is to exponentiate both sides to get $$\frac{(x-1)(x-2)}{x+6}=5^0=1$$ which can of course be rewritten as $$(x-1)(x-2)=x+6$$

So both methods lead to the same result.

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  • $\begingroup$ actually with the rules i mentioned, the x+6 would be being divided. It would be underneath the (x-1)(x-2) as subtraction is equal to division. The textbook moves it to the other side like you did, i dont understand why $\endgroup$ – Saad Siddiqui Mar 10 '16 at 18:11
  • $\begingroup$ That (dividing under the log) would work as well. However, you can choose to just use the rule for two of the terms and then move the $\log_5(x+6)$ to the right hand side. $\endgroup$ – Bobson Dugnutt Mar 10 '16 at 18:13
  • $\begingroup$ If I do the division, then when multiplying x+6 to the other side, it should equal to 0 and thus disappear. This would only leave the left side of my equation equal to zero, This is giving me the wrong answer $\endgroup$ – Saad Siddiqui Mar 10 '16 at 18:16
  • $\begingroup$ @SaadSiddiqui See my edited answer for an explanation on how to solve it using your method. (Note that you exponentiate with $5$ since it is the base the $\log$ is in.) $\endgroup$ – Bobson Dugnutt Mar 10 '16 at 18:19
  • $\begingroup$ Oh! Wow yeah that explains why, when using my method and the quadratic formula, I got the wrong answer. I've never been asked to solve after setting up the equation like that in school, only to set it up. Thank you, that helps a ton! I have a similar question which i'll ask in a bit $\endgroup$ – Saad Siddiqui Mar 10 '16 at 18:26
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Laws of logarithm needed:

$$ \log_a pq = \log_a p + \log_a q$$ $$ \log_a \left(\frac{p}{q}\right) = \log_a p - \log_a q$$

$$\log_b a = 0 \ \implies a = b^0 $$

Using these three laws, your equation can be immediately reduced to:

$$ \log_5 \left(\frac{(x-1)(x-2)}{x+6}\right) = 0$$

Note: this is the step you got it wrong: $$\left(\frac{(x-1)(x-2)}{x+6}\right) = 5^0 = 1 $$

$$ x^2 -3x +2 =x+6$$

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or we have $$\log_5((x-1)(x-2))=\log_5(x+6)$$ thus we get $$(x-1)(x-2)=x+6$$

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  • $\begingroup$ would the x+6 not be being divided? why is it moved to the other side? $\endgroup$ – Saad Siddiqui Mar 10 '16 at 18:12
  • $\begingroup$ @SaadSiddiqui when you remove log from both sides, $0$ on the R.H.S becomes $1$ $\endgroup$ – Nikunj Mar 10 '16 at 20:39
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Well, the mistake in your approach is the use of the $\log$ properties.

Let's rewrite the initial equation as $\log_5\left((x-1)(x-2)\cdot \frac{1}{x+6}\right)=0.$

Then we have (since $\log_5 1=0$ and since $\log$ is $1-1$) that $(x-1)(x-2)\cdot \frac{1}{x+6}=1.$

Then, just do some calculations to arrive at the solution. Do not forget to take the restrictions $x-1>0,$ $x-2>0$ and $x+6>0$ (which are equivalent to $x>2$) in order the $\log_5$ to be well defined.

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  • $\begingroup$ on a previous test, i was able to multiply two log functions with the same base, and then divide by another function of the same base. Something like this: (x-1)(x-2)/(x+6). why does that not work here? $\endgroup$ – Saad Siddiqui Mar 10 '16 at 18:14

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