text: # Bit rusty on my log rearranging
  
  • Jun 12th 2009, 06:18 AM
  Joel
  Bit rusty on my log rearranging
  Can you help me with these
  
  Rewrite these relations in index form (that is, without using logarithims):
  
  (a) logα (x + y) = logα x + logα y
  (b) log ₁₀ x = 3 + log ₁₀ y
  (c) log₃ x = 4log₃y
  (d) 2log₂x + 3log₂y – 4log₂z = 0
  (e) xlogα 2 = log α y
  (f) log α x - log α y = n log α z
  (g) ˝log₂ x + = ⅓log₂ y – 1
  (h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
  • Jun 12th 2009, 06:44 AM
  Amer
  Quote:
  
  Originally Posted by Joel
  Can you help me with these
  
  Rewrite these relations in index form (that is, without using logarithims):
  
  (a) logα (x + y) = logα x + logα y
  (b) log ₁₀ x = 3 + log ₁₀ y
  (c) log₃ x = 4log₃y
  (d) 2log₂x + 3log₂y – 4log₂z = 0
  (e) xlogα 2 = log α y
  (f) log α x - log α y = n log α z
  (g) ˝log₂ x + = ⅓log₂ y – 1
  (h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
  
  $\displaystyle a)log_{a}(x+y) = log_{a} x + log_{a} y$
  
  $\displaystyle log_{a} (x+y) = log_{a} xy$ I use $\displaystyle log A + log B = log (AB)$
  
  $\displaystyle (x+y) = xy$
  
  $\displaystyle b) log_{10} x =3 + log_{10} y$
  
  $\displaystyle x=10^{3+log_{10} y }$
  
  $\displaystyle x= 10^3 (10^{log_{10} y })$
  
  $\displaystyle x= 1000(y)$ since $\displaystyle A^{log_{A} c} = c$
  
  $\displaystyle e)xlog_{a} 2 = log_{a} y$
  
  $\displaystyle log_{a}2^x= log_{a} y$ since $\displaystyle (a) log c = log c^a$
  
  $\displaystyle 2^x = y$
  
  (g) ˝log₂ x + = ⅓log₂ y – 1
  
  $\displaystyle g)1/2log_{2}x = 1/3log_{2} y -1$
  
  $\displaystyle log_{2} x^{1/2} - log_{2} y^{1/3} = -1$
  
  $\displaystyle log_{a} y^{1/3} - log_{2} x^{1/2} = 1$ multiply with -1
  
  $\displaystyle log_{a} \dfrac{y^{1/3}}{x^{1/2}} = 1$
  
  $\displaystyle \dfrac{y^{1/3}}{x^{1/2}} = a$
  • Jun 12th 2009, 06:58 AM
  Amer
  some log rules I attach them because in sometimes latex it is not clear if I use powers
  
  Attachment 11865
  • Jun 12th 2009, 07:43 AM
  Soroban
  Hello, Joel!
  
  Some of your bases didn't show up . . . I'll use $\displaystyle b$.
  
  Quote:
  
  Rewrite in index form (that is, without using logarithims):
  
  $\displaystyle (a)\;\;\log_b(x+y) \:=\:\log_b(x) + \log_b(y)$
  
  We have: .$\displaystyle \log_b(x+y) \:=\:\log_b(xy)$
  
  Then: .$\displaystyle x+y \:=\:xy$
  
  Quote:
  
  $\displaystyle (b)\;\;\log_b(x) \:= \:3 + \log_b(y)$
  We have: .$\displaystyle \log_b(x) \;=\;\log_b(b^3) + \log_b(y) \;=\;\log_b(b^3y)$
  
  Then: .$\displaystyle x \;=\;b^3y$
  
  Quote:
  
  $\displaystyle (c)\;\;\log_b(x) \:=\: 4\log_b(y)$
  We have: .$\displaystyle \log_b(x) \;=\;\log_b(y^4)$
  
  Then: .$\displaystyle x \;=\;y^4$
  
  Quote:
  
  $\displaystyle (d)\;\;2\log_b(x) + 3\log_b(y) - 4\log_b(z) \:=\:0$
  We have: .$\displaystyle \log_b(x^2) + \log_b(y^3) - \log_b(z^4) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{x^2y^3}{z^4}\right) \:=\:0$
  
  Then: .$\displaystyle \frac{x^2y^3}{x^4} \:=\:1$
  
  Quote:
  
  $\displaystyle (e)\;\;x\log_b(2) \:=\: \log_b(y)$
  We have: .$\displaystyle \log_b(2^x) \:=\:\log_b(y)$
  
  Then: .$\displaystyle 2^x\:=\:y$
  
  Quote:
  
  $\displaystyle (f)\;\;\log_b(x) - \log_b(y) \:=\: n\log_b(z)$
  
  We have: .$\displaystyle \log_b(x) - \log_b(y) - \log_b(z^n) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{xy}{z^n}\right) \:=\:0$
  
  Then: .$\displaystyle \frac{xy}{z^n} \;=\;1$
  
  Quote:
  
  $\displaystyle (g)\;\;\tfrac{1}{2}\log_b(x) \:=\: \tfrac{1}{2}\log_b(y) - 1$
  
  We have: .$\displaystyle \log_b\left(x^{\frac{1}{2}}\right) \;=\;\log_b\left(y^{\frac{1}{2}}\right) - \log_b(b)$
  
  . . $\displaystyle \log_b\left(\sqrt{x}\right) - \log_b\left(\sqrt{y}\right) + \log_b(b) \;=\;0 \quad\Rightarrow\quad \log_b\left(\frac{\sqrt{x}}{\sqrt{y}}\cdot b\right) \;=\;0$
  
  Then: .$\displaystyle \frac{b\sqrt{x}}{\sqrt{y}} \:=\:1$
  
  Quote:
  
  $\displaystyle (h)\;\;2\log_b(2x + 1) \:=\: 3\log_b (2x – 1)$
  
  We have: .$\displaystyle \log_b(2x+1)^2 \;=\;\log_b(2x-1)^3$
  
  Then: .$\displaystyle (2x+1)^2 \;=\;(2x-1)^3$
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