Question

Solve: log3 (7x - 6) < log3 (4x + 9)​

Answer 1

$\rm\large\underline{\text{Explanation}}$

First, the logarithm needs to be defined. The power should be positive to define the logarithm.

$\rm\cdots\longrightarrow7x-6>0\text{ and }4x+9>0$

$\rm\cdots\longrightarrow x>\dfrac{6}{7}\text{ and }x>-\dfrac{9}{4}$

$\rm\cdots\longrightarrow x>\dfrac{6}{7}$

The set of solutions that satisfies the inequality is,

$\rm\cdots\longrightarrow7x-6<4x+9$

$\rm\cdots\longrightarrow3x<15$

$\rm\cdots\longrightarrow x<5$

The solution set which is the union of the two is $\dfrac{6}{7}<x<5$.

$\rm\large\underline{\text{Learn more}}$

1. The logarithm consists of three factors which are power, base, and exponent.

$\rm\cdots\longrightarrow a^x=b \iff \log_ab=x$

The relation above also provides that the base cannot be negative, as imaginary values do not exist on the coordinate plane.

Also, the exponential function of base 1 represents a horizontal line. In this case, the logarithm function cannot exist.

The power should always be positive according to that the base is positive.

So, important parts to consider are as follows.

$\rm\cdots\longrightarrow(base)>0\text{ and }(base)\neq1$

$\rm\cdots\longrightarrow(power)>0$

2. The logarithms and the exponential functions are inverse functions of each other.

3. Every logarithm functions go through $(1,0)$ and $\rm((base),0)$.

4. Important formulae of the logarithm.

$\rm\cdots\longrightarrow a^{\log_{a}b}=b$

$\rm\cdots\longrightarrow \log_{a}a=1$

$\rm\cdots\longrightarrow \log_{a}1=0$

$\rm\cdots\longrightarrow log_{a}xy=\log_{a}x+\log_{a}y$

$\rm\cdots\longrightarrow \log_{a}\dfrac{x}{y}=\log_{a}x-\log_{a}y$

$\rm\cdots\longrightarrow \log_{a}{x}^{n}=n\log_{a}x$

$\rm\cdots\longrightarrow \log_{a}b=\dfrac{\log_{c}b}{\log_{c}a}$

$\rm\cdots\longrightarrow \log_{a}b=\dfrac{1}{\log_{b}a}$

$\rm\cdots\longrightarrow \log_{a^{m}}b^{n}=\dfrac{n}{m}\log_{a}b$

$\rm\cdots\longrightarrow a^{\log_{b}c}=c^{\log_{b}a}$