Can you solve 2^x+3^x=13 showing your workings?

Question

btw, the answer is x=2

Answer 1

2^x+3^x=13
since 2 & 3 r +ve
let x be an integer
2^x<=13 & 3^x<=13
so 2^x<=8 & 3^x<=9
so x<=3 & x<=2
so x<=2
also let x be real & non integer
then .......... i m sorry

Answer 2

you can show that 2^x + 3^x is increasing monotnolically

because larger x larger 2^x and lager 3^x

once you put x = 0 LHS = 2
x =1 LHS = 5
x =2 LHS = 13
so x = 2 is a solution and there is no other
'
but had 2^x + 3^x = 12 then x >1 and <2 and one could use some approximation formula

Answer 3

You have yet to prove that this is the only solution
Take
y=2^x+3^x-13
lim y x=>-infinity is -13
lim y x=> + infinity = + infinity
y´= 2^x*ln2 +3^xln3 always>0
so y is increasing from -13 to + infinity and has only one root
By trail you´ll find x=2

Answer 4

Brittany, utilising the "distributive property" you first start up on the parentheses and multiply 4 * 2x and then 4* -3. You get: 8x - 12 = 3x +13 you progression the integers to a minimum of one facet by utilising including 12 to the two factors: 8x - 12 + 12 = 3x + 13 + 12 which provides: 8x = 3x + 25 then you definately progression the x's to a minimum of one facet: 8x - 3x = 3x - 3x + 25 5x = 25 then you definately get the x by utilising itself by utilising dividing the two factors by utilising 5: 5x/5 = 25/5 x = 5 Get it?

Answer 5

2^x + 3^x = 9 + 4

2^x + 3^x = 2^2 + 3^2

x = 2

Answer 6

I stopped using math long ago. You really don't need it with calculators and all. And if I don't have one I always look around for someone to give me the answer. I bet you still think that Latin and French is still spoken.
Dude, you got to party more and relax

Answer 7

Take lg on both sides of the equation,
xlg2+xlg3=lg13
x(lg2+lg3) = lg13
x(lg6) = lg13
x = (lg13)/(lg6) = 1.43

Answer 8

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