Need help with exponents

**ceasar_19134** · August 26th 2008, 12:57 PM

I dont remember how to do this problem and being summer, I dont have my textbook to help.

Solve: 4^(x+2)=9^(x-1)

**Moo** · August 26th 2008, 01:01 PM

Hello !

Originally Posted by ceasar_19134

I dont remember how to do this problem and being summer, I dont have my textbook to help.

Solve: 4^(x+2)=9^(x-1)

Just use logarithms.

Remember this property : $\ln(a^b)=b \ln(a)$

$\ln(4^{x+2})=\ln(9^{x-1})$

$(x+2) \ln(4)=(x-1) \ln(9)$

Can you do the rest ? (you can simplify again in the end by writing $4=2^2 \quad \& \quad 9=3^2$ )

**ceasar_19134** · August 26th 2008, 01:09 PM

I still cant figure it out, sorry. All of the other problems I have to do have the same base or can be simplified to have the same base.

**Moo** · August 26th 2008, 01:22 PM

Originally Posted by ceasar_19134

I still cant figure it out, sorry. All of the other problems I have to do have the same base or can be simplified to have the same base.

Whatever base you choose, the rules are the same

$\ln$ denotes the natural logarithm and is like an *universal* base :
$\ln(e)=1$

Continuing from $(x+2) \ln(4)=(x-1) \ln(9)$ :

$x \ln(4)+2 \ln(4)=x \ln(9)-\ln(9)$

Group the x together :

$x \ln(9)-x \ln(4)=2 \ln(4)+\ln(9)$

$x(\ln(9)-x \ln(4))=2 \ln(4)+\ln(9)$

$\implies x=\frac{2 \ln(4)+\ln(9)}{\ln(9)-\ln(4)} \quad (1)$

Now, if you take the logarithm in base 2, you'll have $\log_2(4)=2$ because $4=2^2$ . Actually, it's $\log_2(4)=\log_2(2^2)=2 \underbrace{\log_2(2)}_{=1}=2$ .

But in a common logarithm, you have :
$\ln(4)=\ln(2^2)=2 \ln(2)$ , same goes for $\ln(9)=\ln(3^2)=2 \ln(3)$

Substituting in (1) :
$x=\frac{4 \ln(2)+2 \ln(3)}{2 \ln(3)-2 \ln(2)}$

Simplify by 2 :

$\boxed{x=\frac{2 \ln(2)+\ln(3)}{\ln(3)-\ln(2)}}$

You gotta learn how to use the natural logarithm. Using bases just sometimes avoid difficulties things easier, but in exercises like this one, it brings difficulties !

**Showcase_22** · August 30th 2008, 05:07 AM

I did it a different way:

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