Finding value of x in logarithms?

Question

Q) Find the value of $x$ in $2 \log x + \log 5 = 2.69897$

So far I got: $$2 \log x + \log 5 = 2.69897$$ $$\Rightarrow \log x^2 + \log 5 = 2.69897 $$

$$\Rightarrow \log 5x^2 = 2.69897 $$

What should I do next?

_{Note: In this question $\log(x) \implies \log_{10}(x)$ , it is therefore implied to use $\ln(x)$ to denote natural logarithm}

Answer 1

I suspect the OP is using base 10 logs instead of the standard base $e$ log. So, using base 10 logs,

We have that $\log(500) = 2.69897...$ and $\log(100)=2$. Thus,

\begin{align*} &\quad 2 \log(x) + \log(5) = \log(500) \\ &\implies 2\log(x) = \log(500) - \log(5) \\ &\implies 2\log(x)=\log(100) \\ &\implies 2 \log(x) = 2 \\ &\implies \log(x) = 1 \\ &\implies x = 10^1 \\ &\implies x=10. \end{align*}

Answer 2

Remember, if $a^x=b$, then $\log_a(b)=x$ $$\log_{10}(5x^2)=2.69897$$ $$10^{2.69897}=5x^2$$ $$x^2=\dfrac{10^{2.69897}}{5}$$ $$x=\pm\sqrt{\dfrac{10^{2.69897}}{5}}$$ We will discard the negative root because we cannot have something like $\log(x)$ where $x < 0$. Therefore: $$\displaystyle \boxed{x=\sqrt{\dfrac{10^{2.69897}}{5}}}$$

Answer 3

Raise the base of the logarithm to both sides. Then, you get $5 x^2 = b^{2.69897}$ where $b$ is the base of the logarithm (probably $b=10$). Then, solve for $x$ by dividing by $5$ and taking square roots.

Answer 4

Oh.

$$2 \log_b x + \log_b 5 = 2.69897=a$$ $$\log_b x=(a-\log_b 5)/2$$ $$x=b^{(a-\log_b 5)/2}$$ And after about a half and a hour you may have $$x=\sqrt{b^{a-\log 5}}=\sqrt{b^a/5}$$ and so on

Answer 5

The comment given on your question is the quickest answer there is.

$\log(5x^2)= \log(500) \implies x=10$

I dislike using a calculator or log table to get the answer. I think such things are supposed to only make lives easier and faster, they should not be the only way to find solutions.

Here's a rough long road just for the sake of independence from machines. Let us consider using a Taylor expansion to estimate the answer. I'll give you two simple expansions that you can use to approximate $\ln(x)$:

$$\ln(1-x) = -\sum\limits_{n=1}^\infty \frac{x^n}{n} \quad\forall\space \mid x\mid\space < 1$$ $$\ln(1+x) = \sum\limits_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n} \quad\forall\space \mid x\mid\space < 1$$

For convenience, let's roll back to the original equation we had: $$2 \log(x) + \log(5) = 2.69897 \\ \Rightarrow 2 \log(x) + \log (0.5\times 10) = 2.69897 \\ \Rightarrow 2 \log(x) + \log (0.5) = 1.69897$$

$$\ln(0.5) = \ln(1-0.5) = -\sum\limits_{n=1}^\infty \frac{(0.5)^n}{n} = -(\frac{(0.5)^1}{1} + \frac{0.5^2}{2} + \frac{0.5^3}{3} + \dots) \approx -0.693147$$

But we want $\log(0.5)$, so we need to use the change of base formula, $$ \log(0.5) = \frac{\ln(0.5)}{\ln(10)} = \frac{-0.693147}{2.302585} \approx -0.301029$$

You can find $\ln(10)$ using the series I showed you and I advise you to keep that value in memory.

Now, $$ 2 \log(x) - 0.301029 = 1.69897 \\ 2 \log(x) = 1.69897 + 0.301029 = 1.999999 \approx 2 \\ \log(x) = 1 \\ x = 10^1$$

---

Some additional points:

• Sometimes you'll be faced to calculate $\text{anti}\log(x)$, for that here's another series:

$$ 10^x = \sum\limits_{n=1}^\infty \frac{x^n \ln^n(10)}{n!}$$

• If you want faster ways of doing it, then please check out Euler's acceleration or if you want really heavy guns, then there's the Cohen-Villegas-Zagier acceleration. But for simple stuff like what we have, Newton-Raphson is good too._{(Courtesy of Balarka Sen)}

• That intermediate base conversion is kind of a bummer for me and most people but remember that you can be creative during such calculation: $\Large \frac{\ln 0.5}{\ln 10} = - \frac{1}{1 + (\ln 5) / (\ln 2)} \approx \frac{1}{1 + (\ln 4) / ( \ln 2 )} = \frac{1}{3}$ _{(Approximation by N3buchadnezzar)}

• Remembering certain values can help. If you don't want to do all of those approximations, then just remember the first 10 or so prime values of $x$ for $\log(x)$ and $\ln(x)$ but I do strongly recommend you remembering $\ln(10)$ and $\log(e)$ becuase base conversions always require those values.
  By remembering a few values, you can find other values simply using your creativity and knowledge of the logarithmic rules: $$ \log(5) = \log(\frac{10}{2}) = \log(10) - \log(2) = 1 - \log(2) \approx 1 - 0.301029 = 0.698971$$