- Thread starter
- #1

#### karush

##### Well-known member

- Jan 31, 2012

- 2,716

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?

- Thread starter karush
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 2,716

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?

- Admin
- #2

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

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?

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

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.

- Thread starter
- #3

- Jan 31, 2012

- 2,716

well that was interesting

why doesn't ln thing work?

why are you up so late???

why doesn't ln thing work?

why are you up so late???

- Admin
- #4

In a nutshell, you applied the rules of exponents/logs incorrectly.well that was interesting

why doesn't ln thing work?

I'm usually up this late.why are you up so late???

- Jan 30, 2018

- 464

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

- Thread starter
- #6

- Jan 31, 2012

- 2,716

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...

$$3^x-14\cdot 3^{-x}=5$$

rewrite

$$3^x-\frac{14}{3^x}=5$$

$\ln$

$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$

so far...

Last edited:

- Admin
- #7

As Country Boy pointed out, you cannot state: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...

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

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

- Jan 30, 2018

- 464

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

Instead Let $y= 3^x$ so that your equation is $y- \frac{14}{y}= 5$. To solve

- Thread starter
- #9

- Jan 31, 2012

- 2,716

Where is the ln(a-b)

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

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

- Jan 30, 2018

- 464

It's right here! You go from $$3^x- \frac{14}{3^x}= 5$$ok let me try the $\ln$ thing... given we have

$$3^x-14\cdot 3^{-x}=5$$

rewrite

$$3^x-\frac{14}{3^x}=5$$

$\ln$

$$\ln3^x-\ln\left(\frac{14}{3^x}\right)=\ln5$$

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.