Finding value of x in logarithms? 
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

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:
$$ \log(5) = \log(\frac{10}{2}) = \log(10) - \log(2) = 1 - \log(2) \approx 1 - 0.301029 = 0.698971$$
