0

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?

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

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?

user477465
  • 720
  • https://proofwiki.org/wiki/Laws_of_Logarithms – lab bhattacharjee May 12 '19 at 18:35
  • 1
    The first equation's solution, at least, is wrong: you cannot do that. Your equation is $$\log\frac{x+2}x=\log(x+2)\implies \frac{x+4}x=x+2\implies \ldots$$ – DonAntonio May 12 '19 at 18:36
  • First step wrong. From $\log_{10}(x+4) -\log_{10}x = \log_{10}(x+2)$ you cannot go to $10^{\log_{10}(x+4)} - 10^{\log_{10}x} = 10^{ \log_{10}(x+2)}$. – GEdgar May 12 '19 at 18:36

3 Answers3

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 $$\log_{3}(x(x^3-8))=\log_{3}(8x)$$

Dr. Sonnhard Graubner
  • 95,283
0

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

$$\log a + \log b = \log(ab)$$ 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.

19aksh
  • 12,768
0

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 $5$.

minori minus
  • 91
  • oh thank you for that clarification. why is it not the case though? because I thought all you have to do it raise it to the power of the base and it will cancel with the log – user477465 May 12 '19 at 18:45
  • It doesn't work because of what I stated above. For your problems, use the fact that $\log a + \log b = \log ab$ and $\log a - \log b = \log {\frac ab}$ – minori minus May 12 '19 at 18:48