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so log(small)ax a^x=x

a=10

Why in this equation does a=10?? I don't understand. In the equation above I understand

*it being x(x-15)=2*just not the a=10 part Please help

- Thread starter goosey00
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- Thread starter
- #1

so log(small)ax a^x=x

a=10

Why in this equation does a=10?? I don't understand. In the equation above I understand

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- Jan 26, 2012

- 4,055

Hi goosey00,

so log(small)ax a^x=x

a=10

Why in this equation does a=10?? I don't understand. In the equation above I understandit being x(x-15)=2just not the a=10 part Please help

Should there be another log? You wrote "log x + (x-15)=2" but I think it might be "log x + log(x-15)=2"

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- Jan 26, 2012

- 4,055

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- Jan 26, 2012

- 4,055

\(\displaystyle \log[x*(x-15)]=2\)

Assuming log x means base 10, then \(\displaystyle 10^2=x(x-15)\)

Can you go further?

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- #8

- Jan 26, 2012

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The way logarithms are defined is \(\displaystyle \log_{a}b=x \implies a^x=b\) When there is nothing written in subscript then we can assume it's 10 (in higher math it could mean "e"). For this problem I think it's safe to say the base is 10. However, you need to remember the definition above in order to switch between the log form and exponential form of an expression.SO, is the rule of a base is 10. You wrote assumed. Thats the confusing part is the 10. I can solve it from there. I just am missing something.

Did your teacher explain what logarithms are and how you can use certain properties to manipulate them?

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- Jan 26, 2012

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The way to learn logs is to make sure you know and understand the definition that I gave you in my last post. It's just a way to rewrite something in an easier form. So be comfortable writing exponents to logs and logs to exponents. Then you'll be introduced to 3-4 rules that only apply to logs and almost all of the problems you'll see use them. Practice, practice, practice.not really but what you just said completely makes sense now. Thanks again. Jenny