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

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.

• Yeah -- "$\log_6$" doesn't mean $\log_{10}6$, it is the name of the logarithm function with base $6$. You're taking the logarithm of the expression $2x+3$, not multiplying it by $\log_{10}6$. Think of $\log_6(x)$ as something like $f(x)$, a function applied to an argument.
– MPW
Commented Oct 4, 2018 at 20:30
• $\log_6$ is not a number. It's like $+$ or $\sqrt{}$ or $\frac {}7$ or $^2$. It's something you do to a number. You can't divide by $\log_6$ any more than you can divide by $+$. Commented Oct 4, 2018 at 20:33
• @fleablood: Note that OP treated it like the base 10 log of 6, "$\log 6$". I confirmed the numerical result.
– MPW
Commented Oct 4, 2018 at 20:34
• Yea, that make sense. Commented Oct 4, 2018 at 20:37

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. $$6^3 = 2x+3$$ $$216 = 2x+3$$ $$213 = 2x$$ $$\boxed{x = \frac{213}{2} = 106.5}$$

• Thanks! This helped me understand logs better! Commented Oct 4, 2018 at 21:08
• No problem, glad to have helped! Commented Oct 4, 2018 at 21:09

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 $$\log_6 (2x + 3) = 3 \iff 6^3 = 2x + 3$$.

• Thank you. This helped a lot! Commented Oct 4, 2018 at 21:08