Rules for equation involving addition and subtraction of logarithms

Question

Given these two examples from my math course:

Example A: $$\log(50) + \log(x/2) = 2 \implies \log 25x = 2 \implies 25x = 10^2 \implies x = 4.$$

Example B: $$\log(72) - \log(2x/3) = 0 \!\implies\! \log(72) = \log(2x/3) \!\implies\! 72 = 2x/3 \!\implies\! 216 = 2x \!\implies \!108 = x$$

My question: Why in Example B, we move $-\log(2x/3)$ over to the other side instead of using the Quotient Property? On the other hand, in Example A we didn't move $\log(x/2)$ to the other side but used the product property instead.

Attempt:

In Example A, if we move $\log(x/2)$ to the other side, we have $$\log(50) + \log(x/2) = 2 \implies \log(50) = -\log(x/2) +2.$$

Now we could raise everything to a common base of $10$ to cancel out the logarithms. So we are left with the following $$50 = -(x/2) + 100.$$

Now we subtract 100 from each side to get

$$-50 = -(x/2) \implies -50 = -(x/2) \implies 100 = 4$$

But as shown in the example $x = 4$.

Why this didn't work?

Answer 1

What you did in the first example was simply not correct. I think I can understand what you meant though, so let’s go through the problem using your way. $$\log (50)+\log \big(\frac{x}{2}\big) = 2$$ $$\log (50) = 2-\log \big(\frac{x}{2}\big)$$ We can rewrite $2$ as a logarithm using base $10$. Use the definition of logs. $$a^x = y \longleftrightarrow \log_a y = x$$ $$10^2 = 100 \longleftrightarrow \log_{10} 100 = 2$$ So, we can use $\log (100)$ instead of $2$.
$$\log (50) = \log (100)-\log \big(\frac{x}{2}\big)$$ Use the Quotient Rule.
$$\log (50) = \log \biggr(\frac{100}{\frac{x}{2}}\biggr)$$ $$\log (50) = \log \big(\frac{100(2)}{x}\big) \implies \log (50) = \log \big(\frac{200}{x}\big)$$ $$50 = \frac{200}{x} \implies x = \frac{200}{50}$$ $$\boxed{x = 4}$$

In the second example, you could have used the Quotient Rule. $$\log (72)-\log \big(\frac{2x}{3}\big) = 0$$ $$\log \biggr(\frac{72}{\frac{2x}{3}}\biggr) = 0 \implies \log \big(\frac{72(3)}{2x}\big) = 0 \implies \log \big(\frac{108}{x}\big) = 0$$ Now, use the definition of logarithms. $$\log_b x = y\longleftrightarrow b^y = x$$ $$\log \big(\frac{108}{x}\big) = 0 \longleftrightarrow 10^0 = \frac{108}{x}$$ $$1 = \frac{108}{x}$$ $$\boxed {x = \frac{108}{1} = 108}$$ As you can see, in both cases, we get the right answer, but they take longer. You have to get used to understanding which ways are “better” than others depending on the type of question.

Answer 2

When you raise the RHS $-\log x/2 + 2$ as the exponent using base 10, you get $10^{-\log x/2 + 2}=10^{-\log x/2}×10^2$ which is $100×(10^{\log x})^{-1/2}=100×(x)^{-1/2}$ The LHS is 50 so $(x)^{-1/2}=1/2$. There is easier way but following your approach. The mistake was when exponentiating, the addition of 2 needed to turn into multiplication by 100.

Answer 3

\begin{eqnarray} \log(50)&=& -\log(x/2) +2\\ \log(50)&=&\log(2/x)+2\\ 50&=&\frac{2}{x}\cdot10^2 \end{eqnarray}

So $x=\frac{200}{50}=4$