How to Solve Exponential Equations with Different Bases?

[nae99]

Homework Statement

4(3^2x+1) + 17(3^x) - 7 = 0

Homework Equations

The Attempt at a Solution

4 + 17 (3^2x+1+x) - 7 = 0

 

[rock.freak667]

I am assuming your problem reads:

4(3^2x+1) + 17(3^x) - 7 = 0

In which case remember that a^m+n=a^maⁿ and a^mn = (a^m)ⁿ.

So after you expand out the first term you can use a substitution to solve for x such as y=3^x.

 

[nae99]

I am assuming your problem reads:

4(3^2x+1) + 17(3^x) - 7 = 0

In which case remember that a^m+n=a^maⁿ and a^mn = (a^m)ⁿ.

So after you expand out the first term you can use a substitution to solve for x such as y=3^x.

and how will i expand it...becaues i don't know how to

 

[rock.freak667]

and how will i expand it...becaues i don't know how to

Take the first term 4(3^2x+1) using the first rule I posted with a^m+n=a^maⁿ, how can you rewrite 3^2x+1 as?

 

[nae99]

Take the first term 4(3^2x+1) using the first rule I posted with a^m+n=a^maⁿ, how can you rewrite 3^2x+1 as?

so is it (3^2x)(3^1)

 

[Ivan92]

Yes. So following up on what rock.freak said, you will have

4(3)(3^2x)+17(3^x)-7=0

So now, you will substitute y=3^x. After that, you should be able to solve it.

 

[nae99]

Yes. So following up on what rock.freak said, you will have

4(3)(3^2x)+17(3^x)-7=0

So now, you will substitute y=3^x. After that, you should be able to solve it.

i have no idea what to do next

 

[SammyS]

3^2x = (3^x)²

& 4(3) = 12

Now, does Ivan's suggestion make more sense to you?

 

[nae99]

3^2x = (3^x)²

& 4(3) = 12

Now, does Ivan's suggestion make more sense to you?

4(3)(3^2x)+17(3x)-7=0

3^2x = (3^x)²

& 4(3) = 12
so after that would it be something like this:

12 + (3^x)^2 + 17(3x)-7=0

 

[Ivan92]

You are not to add 12 but multiply it to the first term. I also noticed that your notation is a bit wrong. Your equation shouldould look like this:

12(3^x)²+17(3^x)-7=0

Then do what rocky said earlier, which is y=3^x. Then you should be able to see what you are to do next.

 

[nae99]

You are not to add 12 but multiply it to the first term. I also noticed that your notation is a bit wrong. Your equation shouldould look like this:

12(3^x)²+17(3^x)-7=0

Then do what rocky said earlier, which is y=3^x. Then you should be able to see what you are to do next.

then
12*17(3^x)-7=0

 

[Ivan92]

No, you would multiply 12 to the (3^x)².

Where ever you see 3^x you substitute in y. When you do that, you should form a quadratic equation. Then do what is necessary to solve it.

12(3^x)²+17(3^x)-7=0

Here is the equation again.

 

[nae99]

No, you would multiply 12 to the (3^x)².

Where ever you see 3^x you substitute in y. When you do that, you should form a quadratic equation. Then do what is necessary to solve it.

12(3^x)²+17(3^x)-7=0

Here is the equation again.

so should it be this
(36^x)² + (51^x)-7=0


which would be
36x^2 + 51x - 7=0

 

[Ivan92]

You just did a common algebraic error:

a(b^x) $\neq$(ab)^x

Just simply put in y for where ever you see 3^x. Then do what I told you before.

 

[nae99]

You just did a common algebraic error:

a(b^x) $\neq$(ab)^x

Just simply put in y for where ever you see 3^x. Then do what I told you before.

12(3^x)²+17(3^x)-7=0

so are you saying i should do this:

12(y)²+17(y)-7=0

 

[SammyS]

No, you would multiply 12 to the (3^x)².

Where ever you see 3^x you substitute in y. When you do that, you should form a quadratic equation. Then do what is necessary to solve it.

12(3^x)²+17(3^x)-7=0

See this equation as corrected by Ivan:

$$12(3^x)^2+17(3^x)-7=0$$

Everyplace you see $3^x$ in the equation, replace it with $y\,.$

You should obtain a quadratic equation, but the variable is y instead of the usual x which you are more accustomed to.

The expression on the left hand side of your new equation can be factored, which should lead to a solution. Alternatively, you should be able to solve it using the quadratic formula.

Added in Edit:

I see you got the quadratic equation while I was entering my post. I'm a slow typist.

 

[nae99]

See this equation as corrected by Ivan:

$$12(3^x)^2+17(3^x)-7=0$$

Everyplace you see $3^x$ in the equation, replace it with $y\,.$

You should obtain a quadratic equation, but the variable is y instead of the usual x which you are more accustomed to.

The expression on the left hand side of your new equation can be factored, which should lead to a solution. Alternatively, you should be able to solve it using the quadratic formula.

ok understand now, so i would be

12y^2+17y-7=0

 

[Ivan92]

Yep. So now all you would have to do now is solve for y. Then when you solve for y, replace y with 3^x then solve for x.

 

[nae99]

Yep. So now all you would have to do now is solve for y. Then when you solve for y, replace y with 3^x then solve for x.

i solve foe y but how should i replace it with 3^x

for eg. y= 1.75
would it then be
3^x=1.75

 

[BloodyFrozen]

Yes, exactly and then use logs to solve it.

 

[nae99]

Yes, exactly and then use logs to solve it.

so is it
x= log3 1.75

 

[SammyS]

i solve foe y but how should i replace it with 3^x

for eg. y= 1.75

Not quite.

I get two solutions for y.

One of them is -7/4 which is equal to -1.75.

What's the other solution?

 

[nae99]

Not quite.

I get two solutions for y.

One of them is -7/4 which is equal to -1.75.

What's the other solution?

the other solution is y=0.333

 

[SammyS]

Actually it's y = 1/3, which is good, because that can be written as 3^-1.

So, all you have to do for this solution to y is to solve: $3^x = 3^{-1}\,.$

Can you solve that for x?

 

[nae99]

Actually it's y = 1/3, which is good, because that can be written as 3^-1.

So, all you have to do for this solution to y is to solve: $3^x = 3^{-1}\,.$

Can you solve that for x?

yes
recall: a^m = a^n, m=n

$3^x = 3^{-1}\,.$
then x= -1