MaxWong

A negative number just means 0 - x . 

e.g. : -1 = 0-1 , - 1/3 = 0 - 1/3.

You can't really 'make' a negative number.

0.545+3[0.545]^1/2

= \(0.545^{\frac{1}{2}}(0.545^{\frac{1}{2}}+3)\)

\(\approx 0.738 (3.738)\\ \text{Which is 2.758644}\)

\(\displaystyle\log_x16 = \dfrac{4}{3}\\ x^{\frac{4}{3}} = 16\\ x^4 = 16^3\\ \;\;\;\;=4096\\ x= \pm\sqrt[4]{4096}\\ \;\;=\pm 8\)

\(\text{log with base -8 is impossible} \therefore \text{x = 8}\)

Same as geno's answer

\(\quad\displaystyle2\log_53 - 3\log_52\\ = \displaystyle\log_5(3^2)-\log_5(2^3)\\ = \displaystyle \log_5(\dfrac{9}{8})\)

\(\boxed {\displaystyle\color{red}a\log b = \log b^a}\\ \color{aqua}and \\\boxed{\displaystyle\color{red}\log a - \log b = \log \dfrac{a}{b}}\)

4¹ = 4

4² = 16

4³ = 64

Therefore 4²ⁿ = XXXXXXXX6     4²ⁿ⁺¹ = XXXXXXXX4

Therefore unit digit of 2354¹⁰⁴⁸ = unit digit of 4¹⁰⁴⁸  = 6.

\(\begin{bmatrix} 1 && 2 \\ 3 && 4 \end{bmatrix}\)

LaTeX code: (\begin{bmatrix}1 && 2  

                     3 && 4 \end{bmatrix}) without the ()

In calculator, Matrices can't be calculated but value of the above one (the 1 2 3 4 one) = 2 x 3 - 1 x 4. 

\(\boxed{\color{red}\text{Dividing a number is same as multiplying its reciprocal}}\)

\(\boxed{\color{olive}\text{Division of a fraction: Exchange the denominator and the numerator and change}\\\color{olive}\div\text{into}\times.} \)

\(\boxed{\color{aqua}\text{If there exists a number x, its reciprocal is}\dfrac{1}{x}}\)

\(\quad\dfrac{1}{6}\div \dfrac{1}{3} \div \dfrac{2}{5}\\=\dfrac{1}{6}\times \dfrac{1}{\dfrac{1}{3}}\times\dfrac{1}{\dfrac{2}{5}}\\ =\dfrac{1}{6}\times 3 \times \dfrac{5}{2}\\=\dfrac{3\times 5}{6\times2}\\=\dfrac{5}{4}\\\color{red}OR\\=1\dfrac{1}{4}\)

x + ln x = 1

ln e^x + ln x = 1

ln xe^x = 1

xe^x = e

Only possible answer is x = 1

Check:

LHS = x + ln x = 1 + ln 1 = 1 + 0 = 1 = RHS.

And please answer there in the original post.

The first line is contradictory. How can a bunch of positive integers add up and became a negative fraction which is \(-\dfrac{1}{12}\)?

But I will solve this.

\(1+2+... = -\dfrac{1}{12}\\ 1^3+2^3+...... = (1+2+.... )^2=(-\dfrac{1}{12})^2=\dfrac{1}{144}\)

This is still contradictory because the sum of a bunch of third powers of a negative number is a positive number.