Math book error? Exponential Equation..

**NotSoBasic** · July 12th 2009, 07:06 PM

$4(2^x)+31(2^x)-8=0$

OK, I have a good feeling it's going to turn into a complex trinomial.
By substituting $(2^x)$ with a variable, say "z"

That would give us..

$4z+31z-8=0$

Still not a trinomial..hmm
I see that the factor may be, -1 and 32.

Or, do I somehow break down the 4 into $2^2$ . But still the exponent 2 would not be over the z.

**Chris L T521** · July 12th 2009, 07:16 PM

Originally Posted by NotSoBasic

$4(2^x)+31(2^x)-8=0$

OK, I have a good feeling it's going to turn into a complex trinomial.
By substituting $(2^x)$ with a variable, say "z"

That would give us..

$4z+31z-8=0$

Still not a trinomial..hmm
I see that the factor may be, -1 and 32.

Or, do I somehow break down the 4 into $2^2$ . But still the exponent 2 would not be over the z.

Well, $4(2^x)+31(2^x)-8=0\implies 35(2^x)=8\implies 2^x=\frac{8}{35}$ .

Now apply logs to both sides and solve for x. Can you continue?

**NotSoBasic** · July 12th 2009, 07:33 PM

Originally Posted by Chris L T521

Well, $4(2^x)+31(2^x)-8=0\implies 35(2^x)=8\implies 2^x=\frac{8}{35}$ .

Now apply logs to both sides and solve for x. Can you continue?

I'll try. Don't think I've had to do this before..

$log2^x=log\frac{8}{35}$

That fraction seems like it cannot be reduced..Unless..

$log2^x=log8-log35$

$log2^x=log2^3-log35$

Hmm..??

**Chris L T521** · July 12th 2009, 07:51 PM

Originally Posted by NotSoBasic

I'll try. Don't think I've had to do this before..

$log2^x=log\frac{8}{35}$

That fraction seems like it cannot be reduced..Unless..

$log2^x=log8-log35$

$log2^x=log2^3-log35$

Hmm..??

You don't need to reduce the fraction.

However, the key to this is to note that $\log 2^{\color{red}x}={\color{red}x}\log 2$ .

Then your equation becomes $x\log 2=\log\frac{8}{35}\implies x=\dots$

Can you finish off?

**NotSoBasic** · July 12th 2009, 08:11 PM

Originally Posted by Chris L T521

You don't need to reduce the fraction.

However, the key to this is to note that $\log 2^{\color{red}x}={\color{red}x}\log 2$ .

Then your equation becomes $x\log 2=\log\frac{8}{35}\implies x=\dots$

Can you finish off?

Oh OK. That is the Law of Logarithms for Powers (LLP)
$logap^n = nlogap$
I've used that before. =]

That fraction is REALLY throwing me off..
$x\log 2=\log\frac{8}{35}$

Next, I think you divide the both sides by log2?
Or do I get the values, like:

$x(.3010)=(-.6409)$
$x=-2.129$

It sorta works, and when subbed into the original equation it equals 0.00156

There must be a better way..

**Chris L T521** · July 12th 2009, 08:34 PM

Originally Posted by NotSoBasic

Oh OK. That is the Law of Logarithms for Powers (LLP)
$logap^n = nlogap$
I've used that before. =]

That fraction is REALLY throwing me off..
$x\log 2=\log\frac{8}{35}$

Next, I think you divide the both sides by log2?
Or do I get the values, like:

$x(.3010)=(-.6409)$
$x=-2.129$

It sorta works, and when subbed into the original equation it equals 0.00156

There must be a better way..

Don't let the fraction get to you!

When you take the log of a fraction, the value is negative.

So it is ok to right the solution as $x=\frac{\log\frac{8}{35}}{\log 2}$ . Now by change of base formula $\log_ab=\frac{\log b}{\log a}$ , we can rewrite our solution as $x=\log_2\left(\frac{8}{35}\right)$ .

Now when you plug it into verify the solution, we keep in mind that $a^{\log_ab}=b$ :

$4\left(2^{\log_2\left(\frac{8}{35}\right)}\right)+ 31\left(2^{\log_2\left(\frac{8}{35}\right)}\right)-8=4\left(\frac{8}{35}\right)+31\left(\frac{8}{35}\ right)-8=35\left(\frac{8}{35}\right)-8=8-8=0$ .

So it shows that $x=\frac{\log\frac{8}{35}}{\log 2}=\log_2\left(\frac{8}{35}\right)$ is the solution to our equation.

**yeongil** · July 12th 2009, 08:57 PM

Originally Posted by NotSoBasic

$4(2^x)+31(2^x)-8=0$

OK, I have a good feeling it's going to turn into a complex trinomial.
By substituting $(2^x)$ with a variable, say "z"

That would give us..

$4z+31z-8=0$

Still not a trinomial..hmm

The problem as given is probably correct, but since you mention trinomials, I have also seen problems where one of the exponents is 2x instead of x, like this:
$2^{{\color{red}2x}} + 6\cdot 2^x + 8 = 0$ .
So, if you substitute $(2^x)$ with another variable (I'll use y), you would get
$y^2 + 6y + 8 = 0$ .

I'd double check wherever you got the problem to make sure that the exponents are correct.

01

**NotSoBasic** · July 13th 2009, 01:02 PM

Originally Posted by Chris L T521

Don't let the fraction get to you!

When you take the log of a fraction, the value is negative.

So it is ok to right the solution as $x=\frac{\log\frac{8}{35}}{\log 2}$ . Now by change of base formula $\log_ab=\frac{\log b}{\log a}$ , we can rewrite our solution as $x=\log_2\left(\frac{8}{35}\right)$ .

Now when you plug it into verify the solution, we keep in mind that $a^{\log_ab}=b$ :

$4\left(2^{\log_2\left(\frac{8}{35}\right)}\right)+ 31\left(2^{\log_2\left(\frac{8}{35}\right)}\right)-8=4\left(\frac{8}{35}\right)+31\left(\frac{8}{35}\ right)-8=35\left(\frac{8}{35}\right)-8=8-8=0$ .

So it shows that $x=\frac{\log\frac{8}{35}}{\log 2}=\log_2\left(\frac{8}{35}\right)$ is the solution to our equation.

That's intense..Good job..

It does appear to work out.
I don't recall doing an equation of this complexity before. Maybe it's one of those "test" questions to make you really scratch your head.

Thanks, I've learned alot.

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