How do you solve a goofy exponent problem using logarithms?

[karush]

$$3^x-14\cdot 3^{-x}=5$$
so expanded
$$x\ln{3}-\ln 7-\ln 2 +x \ln 3=\ln 5$$
ok W|A says the answer is

$$\frac{\ln\left({7}\right)}{\ln\left({3}\right)}$$

don't see the steps how?

 

[MarkFL]

$$3^x-14\cdot 3^{-x}=5$$
so expanded
$$x\ln{3}-\ln 7-\ln 2 +x \ln 3=\ln 5$$
ok W|A says the answer is

$$\frac{\ln\left({7}\right)}{\ln\left({3}\right)}$$

don't see the steps how?

I would first multiply through by \(3^x\) and arrange as:

\(\displaystyle 3^{2x}-5\cdot3^x-14=0\)

We should observe at this point that we have a quadratic in \(3^x\), and we can factor as:

\(\displaystyle \left(3^x-7\right)\left(3^x+2\right)=0\)

Discarding the negative root, we are left with:

\(\displaystyle 3^x=7\)

or:

\(\displaystyle x=\log_3(7)\)

The change of base formula will give us the form cited by W|A.

 

[karush]

well that was interesting

why doesn't ln thing work?

why are you up so late?

 

[MarkFL]

well that was interesting

why doesn't ln thing work?

In a nutshell, you applied the rules of exponents/logs incorrectly.

why are you up so late?

I'm usually up this late. :D

 

[HOI]

log(a- b) is NOT equal to log(a)- log(b) so your first step was wrong.

 

[karush]

ok let me try the $\ln$ thing... given we have
$$3^x-14\cdot 3^{-x}=5$$
rewrite
$$3^x-\frac{14}{3^x}=5$$
$\ln$
$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$
so far...

 

[MarkFL]

ok let me try the $\ln$ thing... given we have
$$3^x-14\cdot 3^{-x}=5$$
rewrite
$$3^x-\frac{14}{3^x}=5$$
$\ln$ it
$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$
so far...

As Country Boy pointed out, you cannot state:

\(\displaystyle \ln(a-b)=\log(a)-\log(b)\)

This isn't an identity, and will give you erroneous results if used.

 

[HOI]

ok let me try the $\ln$ thing... given we have
$$3^x-14\cdot 3^{-x}=5$$
rewrite
$$3^x-\frac{14}{3^x}=5$$
$\ln$
$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$
so far...

No, no, no! This is the same mistake you made before! log(a- b) is NOT log(a)- log(b)!

Instead Let $y= 3^x$ so that your equation is $y- \frac{14}{y}= 5$. To solve that equation. Multiply both sides by y: $y^2- 14= 5y$ which is the same as $y^2- 5y- 14= 0$. Solve that quadratic equation (it factors easily) to get two values for y. Then solve $3^x= y$ for those values of y (if possible) to find x.

 

[karush]

Where is the ln(a-b)

$14\cdot3^{-x}=\frac{14}{3^x}$

 

[HOI]

ok let me try the $\ln$ thing... given we have
$$3^x-14\cdot 3^{-x}=5$$
rewrite
$$3^x-\frac{14}{3^x}=5$$
$\ln$
$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$

It's right here! You go from $$3^x- \frac{14}{3^x}= 5$$
to $$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$.

Taking the logarithm of both sides of $3^x- \frac{14}{3^x}= 5$
would give $ln\left(3^x- \frac{14}{3^x}\right)= ln(5)$
which you then want to write as $ln(3^x)- ln\left(\frac{14}{3^x}\right)= ln(5)$
It is that last step where you have made a mistake.