# Help with ogarithms

#### coooolin32

Well, I'm studying logarithms at the moment, and i have no idea how to do this question Determine the value of x, to 3 d.p.:
3^x = 6^x-1

Help would be much appreciated, and thanks in advance #### coooolin32

Help with Logarithms*

#### undefined

MHF Hall of Honor
Well, I'm studying logarithms at the moment, and i have no idea how to do this question Determine the value of x, to 3 d.p.:
3^x = 6^x-1

Help would be much appreciated, and thanks in advance Do you mean

$$\displaystyle 3^x=6^x-1$$

or

$$\displaystyle 3^x=6^{x-1}$$

?

The second one

#### undefined

MHF Hall of Honor
Write

$$\displaystyle log(3^x)=log(6^{x-1})$$

and use the fact that $$\displaystyle log(n^p)=p\cdot log(n)$$.

#### coooolin32

Yeah i got that, but i have no idea what to do after that.

#### undefined

MHF Hall of Honor
Yeah i got that, but i have no idea what to do after that.
Well it is straightforward solving for x.

$$\displaystyle log(3^x)=log(6^{x-1})$$

$$\displaystyle x\cdot log(3)=(x-1)\cdot log(6)$$

$$\displaystyle x\cdot log(3)=x\cdot log(6) - log(6)$$

$$\displaystyle x\cdot log(3)-x\cdot log(6)= - log(6)$$

$$\displaystyle x(log(3)-log(6))= - log(6)$$

$$\displaystyle x= \frac{-log(6)}{log(3)-log(6)}$$

$$\displaystyle x= \frac{log(6)}{log(6)-log(3)}$$

• coooolin32

#### coooolin32

Thanks, I couldn't get step 3 >.>