Solving logarithmic equations algebraically

Question

1. $\log_{10}(x+4) -\log_{10}x = \log_{10}(x+2)$

$10^{\log_{10}(x+4)} - 10^{\log_{10}x} = 10^{ \log_{10}(x+2)}$

$x+4 - x = x+2$

$x=2$

2. $\ln(x+1)^2 = 2$

$e^{\ln(x+1)^2 } = e^{2}$

$(x+1)^2 = e^2$

$x+1 = \pm e$

$x = -1\pm e$

3. $\ln x +\ln (x^2+1) = 8$

$x +(x^2+1) = e^8 $

can't factor this any further?

4. $\log_{10}8x-\log_{10}(1+\sqrt{x}) =2$

Putting everything to the 10th power gives

$8x-(1+\sqrt x) = 100$

and then solve from there?

5. $\log_3 x +\log_3 (x^2 -8) = \log_3 8x$

$x + x^2 -8= 8x$

$x^2 -7x -8 = 0$

$(x-8)(x+7)=0$, x =8, -7$

The reason i am getting confused is because I don't know where I should use the three rules of logarithms to solve these problems. Do I only use them when i am solving exponential equations?

Answer 1

hint for 1) Write $$\log_{10}\frac{x+4}{x}=\log_{10}{(x+2)}$$ so $$\frac{x+4}{x}=x+2$$ For 3) Write $$\ln(x(x^2+1))=8$$ For 4) It is $$\log_{10}\frac{8x}{1+\sqrt{x}}=2$$ For 5) write $$\log_{3}(x(x^3-8))=\log_{3}(8x)$$

Answer 2

$$\log a - \log b = \log(\frac{a}{b})$$

$$\log a + \log b = \log(ab)$$ So, in your first part,

$$\log_{10}(x+4) - \log_{10}(x) = \log_{10}(x+2)$$

$$\log_{10}\frac{x+4}{x} = \log_{10} (x+2)$$

$$ \frac{x+4}{x} = x+2$$

Similarly it is done in the 3rd, 4th and 5th parts.

Answer 3

This is a comment.

Note that $$10^{\log_{10} {a} + \log_{10} {b}}$$ is not necessarily equal to $$10^{\log_{10}a} + 10^{\log_{10} b}$$

This mistake is made in questions $1, 3, 4$, and $5$.