[SOLVED]6.6.60 goofy exponent prob

karush

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

Staff member
$$3^x-14\cdot 3^{-x}=5$$
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-known member
well that was interesting

why doesn't ln thing work?

why are you up so late???

MarkFL

Staff member
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.

Country Boy

Well-known member
MHB Math Helper
log(a- b) is NOT equal to log(a)- log(b) so your first step was wrong.

karush

Well-known member
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...

Last edited:

MarkFL

Staff member
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.

Country Boy

Well-known member
MHB Math Helper
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

Well-known member
Where is the ln(a-b)

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

Country Boy

Well-known member
MHB Math Helper
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.