User:Rodmjorge/List of logarithmic equations that I learned in 11th grade

This is a list of logarithmic equations that I remember learning in 11th grade, as well as the solution to these equations.

Equation 1

 * $$ \log_2(x) = 3 $$
 * $$ x = 2^3 $$
 * $$ x = 8 $$

Equation 2

 * $$ \log_3(2x)-\log_3(x+2)=0 $$
 * $$ \log_3(2x)=\log_3(x+2) $$
 * $$ 2x=x+2 $$
 * $$ x=2 $$

Equation 3

 * $$ \log_3(\log_2(x))=2 $$
 * $$ \log_2(x)=3^2 $$
 * $$ x=2^{(3^2)} $$
 * $$ x=2^9 $$
 * $$ x=512 $$

Equation 4

 * $$ \log(x)+\log(10x)=3 $$
 * $$ \log(x \times 10x)=3 $$
 * $$ \log(10x^2)=3 $$
 * $$ 10x^2=10^3 $$
 * $$ x^2=10^2 $$
 * $$ x=10 $$

Equation 5
This one was made so you can't put the $$2$$ on the other side.


 * $$ \log_4(x-3)+2=3 $$
 * $$ \log_4(x-3)+\log_4(16)=3 $$
 * $$ \log_4(16(x-3))=3 $$
 * $$ \log_4(16x-48)=3 $$
 * $$ 16x-48=4^3 $$
 * $$ 16x=64+48 $$
 * $$ 16x=112 $$
 * $$ x=7 $$

Equation 6
This one was made so you have to use the logarithm rule about subtractions.


 * $$ \log_3(3x+1)-\log_3(x)=4 $$
 * $$ \log_3(\frac{3x+1}{x})=4 $$
 * $$ \log_3(3+\frac{1}{x})=4 $$
 * $$ 3+\frac{1}{x}=3^4 $$
 * $$ \frac{1}{x}=81-3 $$
 * $$ 1=78x $$
 * $$ x=\frac{1}{78} $$

Equation 7

 * $$ \log_3(x+2)-\log_3(x)=\log_3(2x-1)-\log_3(3x-12) $$
 * $$ \log_3(\frac{x+2}{x})=\log_3(\frac{2x-1}{3x-12}) $$
 * $$ \frac{x+2}{x}=\frac{2x-1}{3x-12} $$
 * $$ (x+2)(3x-12)=(x)(2x-1) $$
 * $$ 3x^2+6x-12x-24=2x^2-x $$
 * $$ 3x^2-6x-24-2x^2+x=0 $$
 * $$ x^2-5x-24=0 $$
 * $$ (x+3)(x-8)=0 $$
 * $$ x=-3 \lor x=8$$

Equation 8

 * $$ \log_3(5+4\log_2(x-1))=2 $$
 * $$ \log_3(5+4\log_2(x-1))=\log_3(9) $$
 * $$ 5+4\log_2(x-1)=9 $$
 * $$ 4\log_2(x-1)=4 $$
 * $$ \log_2(x-1)=1 $$
 * $$ x-1=2^1 $$
 * $$ x=3 $$

Equation 9

 * $$ \log_2(x+2)+\log_2(x-7)=2\log_2(x-4) $$
 * $$ \log_2(x+2)+\log_2(x-7)=\log_2(x-4)^2 $$
 * $$ \log_2((x+2)(x-7))=\log_2(x-4)^2 $$
 * $$ (x+2)(x-7)=(x-4)(x-4) $$
 * $$ x^2+2x-7x-14=x^2-4x-4x+16 $$
 * $$ x^2-5x-14-x^2+8x-16=0 $$
 * $$ 3x-30=0 $$
 * $$ x=10 $$

Equation 10

 * $$ \log_6(\frac{1}{x^2})+\log_6(\frac{1}{x})+\log_6(x)+\log_6(x^2)+\log_6(x^3)=3 $$
 * $$ \log_6(\frac{1}{x^2} \times \frac{1}{x} \times x \times x^2 \times x^3)=3 $$
 * $$ \log_6(x^3)=3 $$
 * $$ x^3=6^3 $$
 * $$ x=6 $$

Equation 11

 * $$ \log_3(2)+2\log_3(x)=\log_3(7x-3) $$
 * $$ \log_3(2)+\log_3(x^2)=\log_3(7x-3) $$
 * $$ \log_3(2x^2)=\log_3(7x-3) $$
 * $$ 2x^2=7x-3 $$
 * $$ 2x^2-7x+3=0 $$
 * $$ (2x-1)(x-3)=0 $$
 * $$ 2x=1 \lor x=3 $$
 * $$ x=\frac{1}{2} \lor x=3 $$

Equation 12
This is just a copy of Equation 2, but you have to use logarithmic subtract rules.


 * $$ \log_3(2x)-\log_3(x+2)=0 $$
 * $$ \log_3(\frac{2x}{x+2})=0 $$
 * $$ \frac{2x}{x+2}=3^0 $$
 * $$ 2x=x+2 $$
 * $$ x=2 $$

Equation 13

 * $$ \log_2(x)+\log_4(x)+\log_16(x)=7 $$
 * $$ \log_2(x)+\frac{\log_2(x)}{\log_2(4)}+\frac{\log_2(x)}{\log_2(16)}=7 $$
 * $$ \log_2(x)+\frac{\log_2(x)}{2}+\frac{\log_2(x)}{4}=7 $$
 * $$ 4\log_2(x)+2\log_2(x)+\log_2(x)=28 $$
 * $$ 7\log_2(x)=28 $$
 * $$ \log_2(x)=4 $$
 * $$ x=2^4 $$
 * $$ x=16 $$

Equation 14

 * $$ \log_3(x)-\log_{\frac{1}{3}}(x^2)=6 $$
 * $$ \log_3(x)-2\log_{\frac{1}{3}}(x)=6 $$
 * $$ \log_3(x)-(-2\log_3(x))=6 $$
 * $$ \log_3(x)+2\log_3(x)=6 $$
 * $$ 3\log_3(x)=6 $$
 * $$ \log_3(x)=2 $$
 * $$ x=3^2 $$
 * $$ x=9 $$

Equation 15

 * $$ \log_7(2)+\log_{49}(x)=\log_{\frac{1}{7}}(\sqrt{3}) $$
 * $$ \log_7(2)+\log_{7^2}(x)=\log_{7^{-1}}(3^{\frac{1}{2}}) $$
 * $$ \log_7(2)+\frac{1}{2} \times \log_7(x)=-\log_7(3^{\frac{1}{2}}) $$
 * $$ \log_7(2)+\frac{1}{2} \times \log_7(x)=-\frac{1}{2} \times \log_7(3) $$
 * $$ 2\log_7(2)+\log_7(x)=-\log_7(3) $$
 * $$ \log_7(4)+\log_7(x)=-\log_7(3) $$
 * $$ \log_7(x)=\log_7(3^{-1})-\log_7(4) $$
 * $$ \log_7(x)=\log_7(\frac{3^{-1}}{4}) $$
 * $$ \log_7(x)=\log_7(\frac{1}{3 \times 4}) $$
 * $$ x=\frac{1}{12} $$

Equation 16

 * $$ \log_{x}(2^{x})=x $$
 * $$ x \times \log_{x}(2)=x $$
 * $$ x \times \log_{x}(2)-x=0 $$
 * $$ x(\log_{x}(2)-1)=0 $$
 * $$ x=0 \lor \log_{x}(2)-1=0 $$
 * $$ x=0 \lor \log_{x}(2)=1 $$
 * $$ x=0 \lor 2=x^1 $$
 * $$ x=0 \lor x=2 $$

Because the defined range does not contain 0, then the solution is $$x=2$$