Solving $\log_{6}(2x+3)=3$. Can I start by dividing by $\log_6$?

Question

For example, $$\log_{6}(2x+3)=3$$

The way I would go about this is solving for $x$.

So we begin by dividing each side by $\log_{6}$:

$$(2x +3) = \frac{3}{\log_{6}}$$

Then subtract $3$:

$$2x = \frac{3}{\log_{6}} -3$$

Then divide each side by $2$:

$$\frac{\frac{3}{\log_{6}} -3}{2}$$

This is equal to $0.428$.

But my math course solves a different way and gets a different answer:

Why did my math course solve in in those specfic steps?

I'm new to logs so please be gentle.

Answer 1

Dividing by $\log_6$ was the mistake. It simply doesn’t make sense, as it has no value.

When we say $\log_b x$, we are referring to an exponent value of $b$ that gives the result $x$. Without the $x$, the statement is meaningless. (I’m guessing you thought it meant $\log_{10} 6$. The base in the question is $6$, not $10$.)

The question itself can be solved simply. Just remember the definition of logarithms. Since logarithms and exponents are inverses of each other, then $$\log_b x = y \longleftrightarrow b^y = x$$ We have the following equation. $$\log_6 (2x+3) = 3$$ Using the definition of logs, we can rearrange this into exponentation form. $$6^3 = 2x+3$$ $$216 = 2x+3$$ $$213 = 2x$$ $$\boxed{x = \frac{213}{2} = 106.5}$$

Answer 2

Okay, you misundertood the question.

You thought it was $(\log 6)\times (2x + 3) = 3$ where $\log 6 = \log_{10} 6$ is the number $k$ where $10^k = 6$.

That is not at all what the problem actually was.

The problem was $\log_6(2x+3) = 3$ where $\log_6 M$ is then number $k$ where $6^k = M$.

So $\log_6(2x+3) = 3$ means $6^3 = 2x + 3$ and ... the rest solves itself.

....

The thing to note is that the $_6$ is in a subscript and that indicates the base. So $\log_b m = k \iff b^k = m$

If you have $\log K$ without a subscript that means that the base is assumed to be $10$. So $\log m = k \iff 10^k = m$.

(TMI: Although more advanced courses often assume $\log$ without a subscript means the base is $e = 2.717....$ so $\log m =k \iff e^k = m$. But that's only more advanced classes that do that. TO play it safe you should always write the base.)

So anyway $\log_6 m = k \iff 6^k =m$. So $\log_6 (2x + 3) = 3 \iff 6^3 = 2x + 3$.