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I need to verify whether or not the below expression is true or not for $z>0$. I'm trying to understand the rules of logarithms but I can't figure out how to apply it myself or where to even begin.

$$\log_5 (z)+\log_{25} (z)= \frac{3}{2}\log_5 (z)$$

Thank you for your help

Lisa
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  • Was the second term supposed to read $\log_{25}(z)$? If so, then the equation is valid. – lulu Apr 11 '16 at 01:09
  • Assuming my interpretation is correct...suppose $\log_5(z)=x$. That means that $5^x=z$. Now suppose that $\log_{25}(z)=y$. That means $25^y=z$. But $25^y=5^{2y}\implies 2y=x\implies y=\frac x2$. Thus $\log_{25} (z)=\frac {log_5(z)}2$ – lulu Apr 11 '16 at 01:14
  • Thanks, fixed! Can you please explain why it is valid? – Lisa Apr 11 '16 at 01:15
  • I just did. Can you follow my comment? – lulu Apr 11 '16 at 01:16
  • Thank you. Is the $^{2y}$ there because $5^2 =25$? Also why does $x$ become $2y$? What happened between $25^y$ and $5^{2y}$ to get that? – Lisa Apr 11 '16 at 01:27
  • Logarithms are simply the inverse of exponents and all the rules apply "in reverse". To get a feel you can raise 5 to each side of the equation to see what happens but that isn't how you should solve it. The trickiest part for this particular problem is to recognize $\log_{25}z = \frac 12\log_5 z$ because $\log_{25}z =c \implies 25^c=z \implies 5^{2c}=z\implies \log_5 z = 2c $. – fleablood Apr 11 '16 at 01:27
  • $25^y=(5^2)^y=5^{2y}$. Since $5^x=25$ and $5^{2y}=25$ we have $5^x=5^{2y}$. It's a very useful property of exponentials that they only take each positive value once. Therefore $5^x=5^{2y}\implies x=2y$. – lulu Apr 11 '16 at 01:29
  • Then just use the rules $k*\log_a z = \log_a z^k $ and $\log_a x + \log_a z = \log_a xz $. Use those three rules and... you can't go wrong. – fleablood Apr 11 '16 at 01:34
  • Thanks I now understand how you got $\frac{log_5(z)}{2}$. What happens with $\frac{3}{2}log_5(z)$ – Lisa Apr 11 '16 at 01:39
  • To your last question: For any $A$ we have $A+\frac A2=A +\frac 12 A=\frac 32 A$. Here we have $\log_5 (z) + \frac 12 \log_5(z)=\frac 32 \log_5 (z)$. – lulu Apr 11 '16 at 01:46

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To elaborate on the comments...

Let's prove: $$\log_{25}(z)=\frac 12 \log_5(z)$$

To see that: suppose $\log_5 (z)=x$. That means $z=5^x$. Now suppose that $\log_{25}(z)=y$. That means $z=25^y$ But $$25^y=5^{2y}=5^x\implies 2y=x \implies y =\frac 12x $$

And that last equality was what we wanted.

Comparison shows that we have established the equality posed in the original question.

lulu
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