solve: 3(4)^(3x-2)=192

Question

jim_thompson5910 · Answer

3(4)^(3x-2)=192

(4)^(3x-2)=192/3

4^(3x-2) = 64

4^(3x-2) = 4^3

3x - 2 = 3 ... the bases are equal (to 4), so the exponents are equal

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x = ??

anonymous · Answer

Yeah so can i marry you?

jim_thompson5910 · Answer

lol no, sry I'm taken, but I'm flattered nonetheless

I'm guessing by that response you found the value of x?

anonymous · Answer

Lol I'm taken too, I just need someone to answer math questions for me. I don't think they needed x value.
Wanna answer another?

jim_thompson5910 · Answer

yes you need to completely isolate x (ie get it by itself)

ah so just marry in the math sense, lol gotcha

jim_thompson5910 · Answer

and sure I can answer another, but tell me what you get for the value of x in this one first

anonymous · Answer

3x-2=3
3x=5
x= 5/3

Next question, $$\log _{2}(x+3)+\log_{2}(x-3)=4$$

jim_thompson5910 · Answer

x = 5/3 is correct for the previous one

jim_thompson5910 · Answer

$$\large \log _{2}(x+3)+\log_{2}(x-3)=4$$

$$\large \log _{2}((x+3)(x-3))=4$$

$$\large (x+3)(x-3)=2^4$$

$$\large (x+3)(x-3)=16$$

$$\large x^2 - 9 =16$$

$$\large x^2 =16+9$$

$$\large x^2 =25$$

so x = ??? or x = ???

make sure you check your two possible answers

anonymous · Answer

okay now i have to marry you.

jim_thompson5910 · Answer

lol I'm sure your boyfriend will be very pleased about that...

jk I know you're not being serious

so what two possible solutions do you get for x?

jim_thompson5910 · Answer

and which possible solution does not work

anonymous · Answer

Positive and Negative 5, i have no clue which one doesnt work

jim_thompson5910 · Answer

plug them in one at a time and see which one is defined and which one is not defined

anonymous · Answer

ohtay, its the negative 5. wellp. i love you.

jim_thompson5910 · Answer

the -5 works or doesn't work?

jim_thompson5910 · Answer

If you're not sure, let's check them one by one

jim_thompson5910 · Answer

Let's check x = 5

$$\large \log _{2}(x+3)+\log_{2}(x-3)=4$$

$$\large \log _{2}(5+3)+\log_{2}(5-3)=4$$

$$\large \log _{2}(8)+\log_{2}(2)=4$$

$$\large \log _{2}(8*2)=4$$

$$\large \log _{2}(16)=4$$

$$\large 4=4$$

so that works, x = 5 is confirmed to be a true solution

jim_thompson5910 · Answer

let's check x = -5

$$\large \log _{2}(x+3)+\log_{2}(x-3)=4$$ 

$$\large \log _{2}(-5+3)+\log_{2}(-5-3)=4$$

$$\large \log _{2}(-2)+\log_{2}(-8)=4$$

this is where we run into trouble, you CANNOT take the log of a negative number...so x = -5 isn't even in the domain ---> x = -5 is NOT a solution

anonymous · Answer

I know, I meant its -5 that doesn't work lol

jim_thompson5910 · Answer

oh gotcha