# Logarithms

#### nuckers

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

$$\displaystyle log3(logx(log416))=-1$$
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

#### undefined

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

$$\displaystyle log3(logx(log416))=-1$$
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.

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

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

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

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

$$\displaystyle x=2^3 = 8$$

• nuckers

#### nuckers

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.

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

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

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

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

$$\displaystyle x=2^3 = 8$$
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.

#### nuckers

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

$$\displaystyle _62log_6x+log_6x=125$$

#### undefined

MHF Hall of Honor
so for this one, i got an answer of x=5, is this correct

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

#### nuckers

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

#### undefined

MHF Hall of Honor
the 2log6x+log6x is all raised, not sure how else to explain it
Is it like this?

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

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

#### nuckers

Is it like this?

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

(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

#### undefined

MHF Hall of Honor
Thats exactly how it's written, good job
$$\displaystyle \displaystyle{6^{2\log_6x+\log_6x}=125}$$

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

$$\displaystyle 3\log_6x=\log_6(125)$$

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

$$\displaystyle x^3=125$$

$$\displaystyle x=5$$

So, to answer your question, yes. • nuckers

#### nuckers

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

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

$$\displaystyle 3\log_6x=\log_6(125)$$

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

$$\displaystyle x^3=125$$

$$\displaystyle x=5$$

So, to answer your question, yes. awesome, exactly what i got, just wanted to double check, some of these logs are so confusing, thanks alot