Negative Solutions for x?

[aisha]

K the question is 3^x^2+20=(1/27)^3x
well I got 3^x^2+20=(3^-3)^3x
the exponents make a quadratic function so i just set it = to 0 and factored to solve for x but this is what i got (x+4)(x+5) this means x=-4, or -5 but doesnt that mean there is no solution? We are not doing Log yet.

[Hurkyl]

(don't forget parentheses when writing expressions)

but doesnt that mean there is no solution?

Okay... why?

[aisha]

Ok i dont know because this question seemed similar to an example problem I had, this A one look below

22x-2x=12 You can convert both sides of the equation to a common base; however, you can solve this exponential equation algebraically (without graphing calculator). Make sure you know how to solve this type of question algebraically.

(2x)2-2x=12 apply law of exponent

Now, if you take a close look, this equation looks like a quadratic equation. Let A be 2x, and rewrite the equation.

A2-A=12

A2-A-12=0

(A+3) (A-4)=0

A= -3 or A= 4

Therefore, 2x= -3, or 2x= 4.

The first equation, 2x= -3, has no solution since any positive base raised to any exponent results in a positive value.

The second equation, 2x=4, gives:
2x =22

Therefore, x = 2.

[aisha]

I did this problem wrong for sure can someone help me out plz :confused:

[dextercioby]

Ok i dont know because this question seemed similar to an example problem I had, this A one look below

22x-2x=12 You can convert both sides of the equation to a common base; however, you can solve this exponential equation algebraically (without graphing calculator). Make sure you know how to solve this type of question algebraically.

(2x)2-2x=12 apply law of exponent

Now, if you take a close look, this equation looks like a quadratic equation. Let A be 2x, and rewrite the equation.

A2-A=12

A2-A-12=0

(A+3) (A-4)=0

A= -3 or A= 4

Therefore, 2x= -3, or 2x= 4.

The first equation, 2x= -3, has no solution since any positive base raised to any exponent results in a positive value.

The second equation, 2x=4, gives:
2x =22

Therefore, x = 2.

I don't mean to be mean,but u have a MAJOR DISSRESPECT FOR THE NOTION OF PARANTHESIS.
When writing down mathematical expressions,paranthesis mean a great deal,since u could get absurd results without taking them into consideration/interpreting the existing ones in a wrong way.
Now,to answer your first post,since the second contained no question,the sollutions u found "x=-4",are and "x=-5" are pefect and more,are unique,and they both check the initial equation as u can easily check.
Note that this time there's noi other equation of the kind "a^{x}=negative number",as in the example posted above.The negative numbers are just the sollution and they needn't be equalled to and exponential function.
Anyway,i find awkward that u study exponential equations before logarithms,as most solutions of such equations can be given in terms of logarithms.

[Nothing]

X being negative does not make it a invalid solution, it's perfectly okay to have negative exponent values, the solutions are -4, -5.

[aisha]

X being negative does not make it a invalid solution, it's perfectly okay to have negative exponent values, the solutions are -4, -5.

Thanks Nothing thats all I wanted to know. :smile:

[aisha]

I don't mean to be mean,but u have a MAJOR DISSRESPECT FOR THE NOTION OF PARANTHESIS.
When writing down mathematical expressions,paranthesis mean a great deal,since u could get absurd results without taking them into consideration/interpreting the existing ones in a wrong way.
Now,to answer your first post,since the second contained no question,the sollutions u found "x=-4",are and "x=-5" are pefect and more,are unique,and they both check the initial equation as u can easily check.
Note that this time there's noi other equation of the kind "a^{x}=negative number",as in the example posted above.The negative numbers are just the sollution and they needn't be equalled to and exponential function.
Anyway,i find awkward that u study exponential equations before logarithms,as most solutions of such equations can be given in terms of logarithms.

YES THANKYOU FOR UR COMMENTS AND HELP :grumpy: