1. ## Logarithms

Not sure how to put this in here, so hopefully i can do my best

$log3(logx(log416))=-1$
The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right

2. Originally Posted by nuckers
Not sure how to put this in here, so hopefully i can do my best

$log3(logx(log416))=-1$
The 3, x and 4 are all the base, so they should be lower than the log and 16. I hope that i said and did that right
Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

$log_3(log_x(log_416))=-1$

$log_x(log_416)=3^{-1}=\frac{1}{3}$

$log_416=x^{\frac{1}{3}}$

$2=x^{\frac{1}{3}}$

$x=2^3 = 8$

3. Originally Posted by undefined
Use underscore, which should be SHIFT plus the key right next to "0" and slightly above and to the right of "P" if your keyboard is like mine.

$log_3(log_x(log_416))=-1$

$log_x(log_416)=3^{-1}=\frac{1}{3}$

$log_416=x^{\frac{1}{3}}$

$2=x^{\frac{1}{3}}$

$x=2^3 = 8$
wow, that was fast and easy, i couldn't remember how the bases work, but i do know, boot the log, thanks so much for your help.

4. so for this one, i got an answer of x=5, is this correct

$_62log_6x+log_6x=125$

5. Originally Posted by nuckers
so for this one, i got an answer of x=5, is this correct

$_62log_6x+log_6x=125$
Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?

6. Originally Posted by undefined
Something strange going on with the 6 and 2 at the beginning. Could you clarify what this is supposed to be?
the 2log6x+log6x is all raised, not sure how else to explain it

7. Originally Posted by nuckers
the 2log6x+log6x is all raised, not sure how else to explain it
Is it like this?

$\displaystyle{6^{2log_6x+log_6x}=125}$

(I know it's kind of hard to read, but if it's right you should be able to tell.)

8. Originally Posted by undefined
Is it like this?

$\displaystyle{6^{2log_6x+log_6x}=125}$

(I know it's kind of hard to read, but if it's right you should be able to tell.)
Thats exactly how it's written, good job

9. Originally Posted by nuckers
Thats exactly how it's written, good job
$\displaystyle{6^{2\log_6x+\log_6x}=125}$

$2\log_6x+\log_6x=\log_6(125)$

$3\log_6x=\log_6(125)$

$\log_6(x^3)=\log_6(125)$

$x^3=125$

$x=5$

10. Originally Posted by undefined
$\displaystyle{6^{2\log_6x+\log_6x}=125}$

$2\log_6x+\log_6x=\log_6(125)$

$3\log_6x=\log_6(125)$

$\log_6(x^3)=\log_6(125)$

$x^3=125$

$x=5$