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[SOLVED] 6.6.60 goofy exponent prob

karush

Well-known member
Jan 31, 2012
2,648
$$3^x-14\cdot 3^{-x}=5$$
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

Administrator
Staff member
Feb 24, 2012
13,734
$$3^x-14\cdot 3^{-x}=5$$
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
Jan 31, 2012
2,648
well that was interesting

why doesn't ln thing work?

why are you up so late???
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,734

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
log(a- b) is NOT equal to log(a)- log(b) so your first step was wrong.
 

karush

Well-known member
Jan 31, 2012
2,648
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

Administrator
Staff member
Feb 24, 2012
13,734
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
Jan 30, 2018
368
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
Jan 31, 2012
2,648
Where is the ln(a-b)

$14\cdot3^{-x}=\frac{14}{3^x}$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
368
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.