Math Help - Logs problem

1. Logs problem

Hi there,

I'm stuck on this logs question and can't find an example in any books that fits this form.

$3^{x} \times 4^{2x+1} = 6^{x+2}$ show that $x = \frac{\ln 9}{\ln 8}$

I try to break the logs down to different forms but never get anything suitable to allow for a substitution.

Any pointers?

Thanks, F

2. Re: Logs problem

Originally Posted by FelixHelix
Hi there,

I'm stuck on this logs question and can't find an example in any books that fits this form.

$3^{x} \times 4^{2x+1} = 6^{x+2}$ show that $x = \frac{\ln 9}{\ln 8}$

I try to break the logs down to different forms but never get anything suitable to allow for a substitution.

Any pointers?

Thanks, F
$3^{x} \times 4^{2x+1} = 6^{x+2}$

Taking logs:

$ln(3^{x}4^{2x+1})=ln(6^{x+2})$

Splitting logs:

$ln(3^x)+ln(4^{2x+1})=ln(6^{x+2})$

Using the power rule on each individual term:

$xln(3)+(2x+1)ln4=(x+2)ln6$

Isolating the x's:

$xln(3)+2xln(4)-xln(6)=2ln(6)-ln(4)$

Taking out a factor of x and rewriting things using the power rule:

$x(ln(3)+2ln(4)-ln(6))=ln(36)-ln(4)$

See if you can proceed, using the rules that:

$log(ab)=log(a)+log(b)$

$log(\frac{a}{b})=log(a)-log(b)$ and

$log(x^n)=nlog(x)$

3. Re: Logs problem

I would start by first rewriting $4^{2x+1}$ and $6^{x+2}$ into $4^{2x} \times 4^1$ and $6^x \times 6^2$ respectively. Then go from there. See if that is enough to get you going.

4. Re: Logs problem

Doing it by using logs like Quacky suggested is one way. What I suggested leads to:

$3^x \times 4^{2x+1} = 6^{x+2}$

$3^x \times 4^{2x} \times 4^1 = 6^x \times 6^2$

Rearrange terms and get:

$\frac{3^x \times (4^2)^x}{6^x} = \frac{6^2}{4}$

Simplify:

$(\frac{3 \times 16^}{6})^x = 9$

$8^x = 9$

From here, take the natural log of both sides and "Wallah!"

5. Re: Logs problem

Thanks everyone - I'd gone down that avenue of rewriting but must have made a mistake (several times!)

Thanks again.

F