Outline to solve: Multiply by the conjugate of any term in LHS. It will yield a quadratic equation in term of the conjugate. Take the logarithm of the root. Let u=[4+sqrt(15)]^x v=[4-sqrt(15)]^x Then u+v=62 Multiply by v, say: 1+v²=62v --> v²-62v+1=0 v=½[62±sqrt(62²-4)] =½[62±sqrt(3840)] =½[62±16sqrt(15)] =31±8sqrt(15) x=[log(v)]/log(4-sqrt(15) =4±[1/log{4-sqrt(15)}]
Use of conjugate applications will be a clue. 4 - √15 = 1/4 +√15 . On substitution it gives you m^2 - 62 m+1=0 where m= (4+√15) ^x = m, solving the quadratic m = (31 + - 8√15) . recall m = (4 + √15) ^x take log and get the value of x
Be f: x ---> (4 + sqrt(15))^x + (4 - sqrt(15))^x. f(x) = (4 + sqrt(15))^x + (4 + sqrt(15))^(-x) = exp(x.ln(4 + sqrt(15))) + exp(-x.ln(4 + sqrt(15))) = 2.cosh(x.ln(4 + sqrt(15))) f verifies f(x) = f(-x) for any real x, and if we limit f at R+ then it is a bijection from R+ to [2, + infinity[ As 65 is in [2, +infinity], the equation f(x) = 65 has an unique solution x0 on R+ and an unique solution -x0 on R- As x0 = 2 is evident solution we then have that the given equation has two solutions and only two: 2 and -2.
X=2......May be ^=read as to the power *=read as square root Let R=4+*15 A=4-*15 RA=(4+*15)(4-*16) =4^2-(*15)^2 =16-15=1 So, RA=1 A=1/R....... Eqn1 Let R^x=t As per question t+(1/t)=62 So, (t^2+1)/t =62 t^2+1=62t t^2-62t+1=0 Here a=1,b=(-62),c=1 D=b^2-4ac =(-62)^2 -{4×1×1) =3844-4=3840 *D=*(3840) =(*256)×(*15) =16.*15 So, t={-(-62)±16.*15}/2 =(62+16.*15)/2 =2(31+8.*15)/2 =31+8.*15 =4^2+(*15)^2+(2×4×*15) =(4+*15)^2 =R^2 So, t=R^2 R^x=R^2...(t=R^x) So, X=2 We can also apply logarithmic method...... R^x=R^2 Take log logR^x=logR^2 X. logR=2.logR X=2.logR/logR X=2
Outline to solve:
Multiply by the conjugate of any term in LHS. It will yield a quadratic equation in term of the conjugate. Take the logarithm of the root.
Let u=[4+sqrt(15)]^x
v=[4-sqrt(15)]^x
Then u+v=62
Multiply by v, say:
1+v²=62v --> v²-62v+1=0
v=½[62±sqrt(62²-4)]
=½[62±sqrt(3840)]
=½[62±16sqrt(15)]
=31±8sqrt(15)
x=[log(v)]/log(4-sqrt(15)
=4±[1/log{4-sqrt(15)}]
Use of conjugate applications will be a clue. 4 - √15 = 1/4 +√15 . On substitution it gives you m^2 - 62 m+1=0 where m= (4+√15) ^x = m, solving the quadratic m = (31 + - 8√15) . recall m = (4 + √15) ^x take log and get the value of x
Nice solution and good explanation
Thank you
X has to be an even integer because if it were odd there would be a cross term involving the square root of 15..Testing 2 yields the result .
(2^2+2^3)+(2^2+2^3) (1^1+1^1)+(2^1+1^3) (2+3)(x ➖ 3x+2).
Be f: x ---> (4 + sqrt(15))^x + (4 - sqrt(15))^x.
f(x) = (4 + sqrt(15))^x + (4 + sqrt(15))^(-x) = exp(x.ln(4 + sqrt(15))) + exp(-x.ln(4 + sqrt(15))) = 2.cosh(x.ln(4 + sqrt(15)))
f verifies f(x) = f(-x) for any real x, and if we limit f at R+ then it is a bijection from R+ to [2, + infinity[
As 65 is in [2, +infinity], the equation f(x) = 65 has an unique solution x0 on R+ and an unique solution -x0 on R-
As x0 = 2 is evident solution we then have that the given equation has two solutions and only two: 2 and -2.
X=2......May be
^=read as to the power
*=read as square root
Let R=4+*15
A=4-*15
RA=(4+*15)(4-*16)
=4^2-(*15)^2
=16-15=1
So,
RA=1
A=1/R....... Eqn1
Let R^x=t
As per question
t+(1/t)=62
So,
(t^2+1)/t =62
t^2+1=62t
t^2-62t+1=0
Here a=1,b=(-62),c=1
D=b^2-4ac
=(-62)^2 -{4×1×1)
=3844-4=3840
*D=*(3840)
=(*256)×(*15)
=16.*15
So,
t={-(-62)±16.*15}/2
=(62+16.*15)/2
=2(31+8.*15)/2
=31+8.*15
=4^2+(*15)^2+(2×4×*15)
=(4+*15)^2
=R^2
So,
t=R^2
R^x=R^2...(t=R^x)
So,
X=2
We can also apply logarithmic method......
R^x=R^2
Take log
logR^x=logR^2
X. logR=2.logR
X=2.logR/logR
X=2
Solution by insight
(a+b)^2+(a-b)^2=
2(a^2+b^2)=
2(16+15)=62
x=2
And. a+b=1/(a-b)
x=-2 is also an answer
Very very very slow explanation boring
लगता है तुम्हे म्यथ्स सब्जेक्ट ही बोर है।
The walk of mathematics is adventurous. So, Sir, You shouldn't present opinion like this.