# Thread: [SOLVED] Log true or false?

1. ## [SOLVED] Log true or false?

True or False?
For positive real numbers a and b,
$\displaystyle \frac {\log_{10} b}a = \log_{10} b^{1/a}$

I think its false but I don't know how to show that it is.
Can someone please tell me how to put fractions on this forum.
Thanks!

2. Originally Posted by yoman360
True or False?
For positive real numbers a and b,
$\displaystyle log_{10} b/a = log_{10} (b^{1/a})$

I think its false but I don't know how to show that it is.
Can someone please tell me how to put fractions on this forum.
Thanks!
do you mean $\displaystyle \frac {\log_{10} b}a = \log_{10} b^{1/a}$ or $\displaystyle \log_{10} \left( \frac ba \right) = \log_{10} b^{1/a}$

the first case is true, the last is false. do you see why?

3. Originally Posted by Jhevon
do you mean $\displaystyle \frac {\log_{10} b}a = \log_{10} b^{1/a}$ or $\displaystyle \log_{10} \left( \frac ba \right) = \log_{10} b^{1/a}$

the first case is true, the last is false. do you see why?
Yeah its the first one.

4. Originally Posted by yoman360
Yeah its the first one.
ok, then there is just one step. recall that $\displaystyle \log_a x^n = n \log_a x$

5. Originally Posted by Jhevon
ok, then there is just one step. recall that $\displaystyle \log_a x^n = n \log_a x$
is this correct?
$\displaystyle \frac {\log_{10} b}a = \log_{10} b^{1/a}$
<-------------- correct?

6. Originally Posted by yoman360
is this correct?
$\displaystyle \frac {\log_{10} b}a = \log_{10} b^{1/a}$
<-------------- correct?
don't know what you were trying to say for the first part, but the last three lines are correct, that's all you need to show

7. Originally Posted by Jhevon
don't know what you were trying to say for the first part, but the last three lines are correct, that's all you need to show
well using the change of base formula:
$\displaystyle log_{10} b = (log_{10} b)/log_{10} 10$
$\displaystyle log_{10} 10=1$ so...
$\displaystyle (log_{10} b)/1= log_{10} b$

how do you put fractions in this place?

8. I think his/her last posting is about change of base, and also some confusion about notation.
More and more, we see this being used: $\displaystyle \log (x)\text{ for }\ln (x)$.
Most computer algebra systems use that convention.
Nevertheless, historically $\displaystyle \log (x)\text{ meant }\log_{10} (x)$ hence confusion.

Using the newer notation we get the classic change of base formula.
$\displaystyle \log _{10} (b) = \frac{{\log (b)}}{{\log (10)}}$.
In this usage it is the case that $\displaystyle \log (10)\not=1$.

9. Originally Posted by Plato
I think his/her last posting is about change of base, and also some confusion about notation.
More and more, we see this being used: $\displaystyle \log (x)\text{ for }\ln (x)$.
Most computer algebra systems use that convention.
Nevertheless, historically $\displaystyle \log (x)\text{ meant }\log_{10} (x)$ hence confusion.

Using the newer notation we get the classic change of base formula.
$\displaystyle \log _{10} (b) = \frac{{\log (b)}}{{\log (10)}}$.
In this usage it is the case that $\displaystyle \log (10)\not=1$.
$\displaystyle log_{10} 10= 1$ Read it like this 10 to what power equals 10. The answer is 1.

or

$\displaystyle log_{10} 10= x$ rewrite in exponential form

$\displaystyle 10^x= 10$

x= 1 so $\displaystyle log_{10} 10= 1$

Get it?

10. log_{10} x = log x
change of base formula:
$\displaystyle log_x y = \frac{{\log_{10} (y)}}{{\log_{10} (x)}}$

or
$\displaystyle log_x y = \frac{{\log (y)}}{{\log (x)}}$

11. Originally Posted by yoman360
Get it?
What do you mean by that?
I simply pointed out to you that you may be confused as to how $\displaystyle \log$ notation is being used in your text material.
Although in most modern texts and software $\displaystyle \log (x)\text{ is used for the older }\ln (x)$.