If (Sqrt3)^5 * (9^2) = (3^x)*(3*Sqrt3) Then x equals :

Arun Maharana   8 years AGO

sqrt3^5 *9^2=(3^x)*(3*sqrt3)
=>3^(5/2)*3^4=(3^x) * 3^(3/2)
=>3^((5/2)+4)=3^(x+3/2)
=>x+(3/2)=13/2
=>x=10/2
=>x=5

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Shagun Chaudhary   8 years AGO

equating the power of 3 and sqrt 3, we get x+1=6
x=5

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MRITUNJAY   8 years AGO

sqrt3^4*sqrt3*3^4=3^x*3*sqrt3
3^2*3^4*sqrt3=3^x*sqrt3
after solving we get x=5
option 4

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ritu yadav   8 years AGO

(sqrt3)^5*9^2= 3^x*3(sqrt3)
(sqrt3)^5*(sqrt3)^8=(sqrt3)^2x*(sqrt3)^3
equating the powers of sqrt3
we get,
5+8=2x+3
x=5

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Prakash Panda   8 years AGO

(sqrt3)^5*(9^2)=(3^x)*(3*sqrt3)
==>(3^x)=((sqrt3)^5*(9^2))/(3*sqrt3)

we know that (sqrt 3)^5= 9*(sqrt 3)
so that part cancled n remain 3

now (3^x)=3*(9^2)

by solving dis we get the ans as option 4 i.e 5

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Abhineet kumar   8 years AGO

(?3)^5*9^2=3^x*(3?3)
3*3?3*3^4=3^x*3?3
3^5=3^x
so
x=5

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suryakant singh   8 years AGO

answer: x=5
EXPALNATION:
(sqrt3)^5 = 3^5/2
(3*sqrt3)=3^3/2
if take log in both side then
log(3^5/2*9^2) = log(3^x*(3^3/2))
log(3^5/2)+log(9^2)=log(3^x)+log(3^3/2)
5/2log(3)+4log(3 )=xlog(3)+3/2log(3)
if we take log(3)common and cancelled then
5/2+4 = x+3/2
then x=5

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PRAGATHI G   8 years AGO

by power law..
3^((5/2)+4) = 3^(x+(3/2))
cancelling the bases and equating the powers
(5+8)/2 = (2x+3)/2
x = 5

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Sagar Jinachandran   8 years AGO

Left Hand Side
=Sqrt 3= 1.732
=(sqrt3)^5=15.58
=(sqrt(3)^5)*(9^2)=1262.665
RHS
=3^x*(3*sqrt(3))
=3^x*(3*1.732)
=3^x*5.196152
Comparing LHS and RHS and solving
we get
1262.665=(3^x*5.196152)
or 3^x=243
solving we get x=5
option 4

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Abhishek   8 years AGO

(Sqrt3)^5* 9^2 = (3^x) * (3*Sqrt3)

On simplifying:
3^(5/2) * 3^4 = 3^x * 3^(3/2)

on further simplifying:
3^(13/2) = 3^((2x+3)/2)

13/2 = (2x+3)/2
13 = 2x+3
x = 5

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Anurag Bawankar   8 years AGO

wrting each term in power of 3 we get

(3)^6.5 =(3)^x+1.5

x+1.5=6.5

x=5

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sachin sawant   8 years AGO

3^(5/2)*3^4=3^( x+3/2)
5/2+4=x+3/2
x=5

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GOURANGA SUNDAR MANDAL   8 years AGO

(sqrt3)^5*(9^2)
=(sqrt3)^13.....equ 1
(3^x)*(3*sqrt3)
=(sqrt3)^3+2x...equ 2
..calculating this two equ..we get x=5..
so option 4 is correct answer

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suprita hadimani   8 years AGO

(sqrt 3)^5*(9^2)=(3^x)*(3*sqrt3)
1.732^5 * 81= (3^x)*(5.196)
15.5861*81=(3^x)*(5.196)
1262.27/5.196=3^x
242.96=3^x
x=5

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kriti chaudhary   8 years AGO

L.H.S
(sqrt3)^5*9^2=(sqrt3)^5*9^2
=(sqrt3)^2(sqrt3)^2(sqrt3)*9^2=3*3*(sqrt3)*9^2=
9*(sqrt3)*9^2=(sqrt3)*9^3=729*(sqrt3)
R.H.S
(3^x)*(3*sqrt3)=(3^x+1)*(sqrt3)
if x=5,then 3^5=243 and x+1=6,then 3^6=729
so,3^6*(sqrt3)=729*(sqrt3)
L.H.S=R.H.S
So value of x=5.
answer.x=5

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Jamil Ahmed   8 years AGO

(sqrt3^5)*(9^2) = (3^x)*(3*sqrt3)
or, [3^(5/2)]*[(3^2)^2] = (3^x)*[3^(3/2)]
or, 3^(5/2+4) = 3^(x+3/2)
or, 5/2 + 4 = x + 3/2
or, x = 5/2 + 4 - 3/2 = 5

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SARANYA   8 years AGO

(SQRT3)^5*(9^2) = 3^X*(3*SQRT3)
(SQRT3)^5*(3^4) = 3^X*(3*SQRT3)
(SQRT3)^2*3^4 = 3^X
3*3^4 = 3^X
X=5

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Rajanikanta Pradhan   8 years AGO

As (sqrt3)^5 *(9^2)=(3^x)*(3*sqrt3)
so this can be rewritten as
fallows (sqrt3)*((sqrt3)^4) *((3^2)^2)=((3)^x)* 3*(sqrt3)
do by solving this we can get x=5

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sumit kumar   8 years AGO

(?3)^5*9^2=3^x*3?3
3^5/2*(3^2)^2=3^x*3^3/2
3^(5/2+4)=3^(3/2+x)
3^13/2=3^3/2+x
13/2=3/2+x
x=13/2-3/2=10/2=5
option 4

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Neelima Kongathi   8 years AGO

(sqrt3)^5 * (9^2) = (3^x)* 3 (sqrt3)
(sqrt3)^4 * 3^4 = 3^(x+1)
3^6 = 3^(x+1)
if bases are equal then powers are also equal
6 = x+1
x = 5

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Venkateswarlu   8 years AGO

3^(5/2)*9^2=3^(x)*3*3^(1/2);
3^(2)*3^(4)=3^(x+1);
3^(6)=3^(x+1);
x+1=6;
x=5;

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Laxmi Venkata Mownika Kurapati   8 years AGO

(sqrt3)^4*(3^4)=(3^x)*3
(3^4)*(3)=(3^x)
3^6=3^x
x=6

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Devendra Marghade   8 years AGO

(Sqrt3)^5*9^2 =(3^x)*(3*Sqrt3)
(Sqrt3)^5*(Sqrt3)^8=(Sqrt3)^2x*(Sqrt3)^3
(Sqrt3)^13 =(Sqrt3)^(2x+3)
So 13=2x+3
x=10/2=5
Option 4)

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vicky mehta   8 years AGO

3*sqrt3*9*9=3^x*3sqrt3
Then
3*3*3*3*3=3^x
Hence
X=5

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Shikhar Awasthi   8 years AGO

(√3)^5*9^2=3^x*3(√3)
(√3)^4*9^2=3^x*3
9*9*9=3^x*3
3^5=3^x
By comparing
X=5

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Vibhay Thakur   8 years AGO

--(Sqrt3)^5*(9^2)=(3^x)*(3*Sqrt3)

--(Sqrt3)*(Sqrt3)*(Sqrt3)*(Sqrt3)*(Sqrt3)*(9^2)=(3^x)*(3*Sqrt3)

--(Sqrt3)*(Sqrt3)*(Sqrt3)*(Sqrt3)*(9^2)=(3^x)*(3)

--9*(9^2)=(3^x)*(3)

--3*(81)=3^x

--3*3*3*3*3=3^x

--3^5=3^x
--x=5

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Ankur Chaurasia   8 years AGO

3^(5/2) * 3^4 = 3^x * 3^(3/2)
hence 13/2 = (2*x+3)/2
therefore 2*x+3 = 13
so x=5

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Mrs.Thangaraj   8 years AGO

(3^1/2)^5 * (3^2)^2=3^x*3^1*3^1/2
Apply the rule (a^x)^y = a^xy
3^5/2*3^4=3^x*3^1*3^1/2
Apply the rule a^x*a^y=a^(x+y)
3^(5/2+4)=3^(x+1+1/2)
5/2+4=x+1+1/2
(5+8)/2=(2x+2+1)/2
13=2x+3
2x=10
X=5

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Parag Wagh   8 years AGO

(3*sqrt3)*3*3*3*3*3=(3^x)*(3*sqrt3)
3^5 = 3^x
x = 5

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karnatakam.Jaswanth kumar   8 years AGO

Split the square roots and compare the square root terms and non sqaue root terms on both sides

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Rishu   8 years AGO

((sqrt3)^5)*(9^2)=(3^x)*3*(sqrt3)
((sqrt3)^4)*(9^2)=3^(x+1)
9*(9^2)=3^(x+1)
9^3=3^(x+1)
3^6=3^(x+1)
since bases are same
6=x+1
x=5

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monu kumar   8 years AGO

3^(5/2+4)
=3^(x+1+/2)
SO x=5

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divya   8 years AGO

by solving the given equation we get
3^6=3^x*3
so 6=x+1
x=5

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Saleel Kulkarni   8 years AGO

3^5/2*3^4=3^x*3^3/2
3^5=3^x
Taking logarithm on both sides
5log3=xlog3
i.e. x=5
Option 4

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Vijay H   8 years AGO

convert to 3 powers, and equate the powers and solve for x.
equated equation will be (5/2)+4 = x+(3/2)

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Behara Avinash   8 years AGO

All the bases should be converted into 3 and if bases all are equal power should be added then we get answer as 5.

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Gaurab Kumar Jha   8 years AGO

A/Q 9*sqrt 3 * 81 = 3^x * 3 * sqrt 3
81 * 3 = 3 ^ x
3 ^ 5 = 3 ^ x
therefore x = 5 i,e option 4 .

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Pradeepta Chattopadhyay   8 years AGO

9=(sqrt3)^4
so LHS = (sqrt3)^13
RHS=(sqrt3)^(3+2x)
therefore, 3+2x=13
thus x=5...

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