Exponential equations grade 11 math

[From: ] [author: ] [Date: 12-03-20] [Hit: ]

5783637589624105473245267Next, ask him to add all the number and he gets 257834Show him the number which you had written as an answer to this addition earlier in the folded paper and surprise him.-a)3^2x=5*(3^x) + (3^2 * 2^2)=3^2x/3^x= 5 + (3^2 * 2^2)/3^x=2x/x=2, therefore 3^2x/3^x=3^2=99=5+(3^2 * 2^2)/3^x4=3^2-x*2^21=3^2-x1=(3^0)^2-xtherefore no solution since 1=1 was the end result.......

a sheet of paper and keep it folded. You have to write the
number by subtracting 2 from the above number and
adding 2 in front which will be....
257834
Next, ask the person to write another five-digit number
below his original number. Suppose he writes 37589.
So you now have...
57836
37589
Now you write a five-digit number below it in such a
way that each digit is 9 minus digit above. You now have...
57836
37589
62410
Ask the person to add one more 5-digit number. If he
adds 54732, you add below it 45267. Note that you decided
the number by subtracting each of his digit from 9.
Thus, you now have.....
57836
37589
62410
54732
45267
Next, ask him to add all the number and he gets 257834
Show him the number which you had written as an answer
to this addition earlier in the folded paper and surprise him.

-

a)3^2x=5*(3^x) + (3^2 * 2^2)
=3^2x/3^x= 5 + (3^2 * 2^2)/3^x
=2x/x=2, therefore 3^2x/3^x=3^2=9
9=5+(3^2 * 2^2)/3^x
4=3^2-x*2^2
1=3^2-x
1=(3^0)^2-x
therefore no solution since 1=1 was the end result. (I'm not sure about the first question so you might have to ask the teacher on that one)
b)(2^-3)^x-3=2*(2^4)^2x+1
=2^-3x+9=2*2^8x+4
=2^-3x+9=2^8x+5
=-3x+9=8x+5
=-11x=14
x=-14/11

c)3^x^2+20=(3^-3)^3x
=3^x^2+20=3^-9x
=x^2+20=-9x
x^2+9x+20=0
(x+4)(x+5)=0
x=-4 or x=-5

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