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ANY INTEGER IS DENOTED BY N
IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER.
IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS
2N+1 IS ODD INTEGER.
SUM OF ODD AND EVEN INTEGERS IS
(2N+1)+(2M)=2N+2M+1=2(N+M)+1=2P+1 WHERE P IS AN INTEGER.
HENCE BY THE ABOVE LOGIC THE SUM IS AN ODD INTEGER.

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ANY INTEGER IS DENOTED BY N
IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER.
PRODUCT OF 2 EVEN INTEGERS IS
(2N)(2M)=4MN=2(2MN)=2P WHERE P IS AN INTEGER.
HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN EVEN INTEGER.

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ANY INTEGER IS DENOTED BY N
IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER.
IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS
2N+1 IS ODD INTEGER.
PRODUCT OF 1 ODD AND 1 EVEN INTEGER IS
(2N)(2M+1)=4MN+2N=2(2MN+N)=2P WHERE P IS AN INTEGER.
HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN EVEN INTEGER.

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ANY INTEGER IS DENOTED BY N
IT WILL BE EVEN IF WE MULTIPLY BY 2...THAT IS 2N IS EVEN INTEGER.
IT WILL BE ODD IF WE ADD 1 TO IT..THAT IS
2N+1 IS ODD INTEGER.
PRODUCT OF 2 ODD INTEGERS IS
(2N+1)(2M+1)=4MN+2N+2M+1=2(2MN+M+N)+1=2P+1 WHERE PIS AN INTEGER.
HENCE BY THE ABOVE LOGIC THE PRODUCT IS AN ODD INTEGER.

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g(x)=x-3
solve if g[g(6)]
G(6)=6-3=3...(TO FIND G(A)..PUT X=A IN THE GIVEN EQN. FOR G(X)
SO NOW G(G(6))=G(3)=3-3=0

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2x-3y=6
3y=2x-6
y=2x/3-2
slope=2/3
y intercept point is (0,2)
see the following example and do the same way..
-------------------------------------------------
Slope: ; -3/4 y-intercept: (0, 8)
eqn.of a line with slope..m ..and y intercept..c...is given by
y=mx+c...hence..eqn.of reqd.line is
y=(-3x/4)+8..or....4y=-3x+32
determine points and graph as follows
x....................0.................4.....................8.....etc
y=(-3x/4)+8..........8.................5.....................2...etc..

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see the following example and do the same way..
-------------------------------------------------
Slope: ; -3/4 y-intercept: (0, 8)
eqn.of a line with slope..m ..and y intercept..c...is given by
y=mx+c...hence..eqn.of reqd.line is
y=(-3x/4)+8..or....4y=-3x+32
determine points and graph as follows
x....................0.................4.....................8.....etc
y=(-3x/4)+8..........8.................5.....................2...etc..

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Slope: ; -3/4 y-intercept: (0, 8)
eqn.of a line with slope..m ..and y intercept..c...is given by
y=mx+c...hence..eqn.of reqd.line is
y=(-3x/4)+8..or....4y=-3x+32
determine points and graph as follows
x....................0.................4.....................8.....etc
y=(-3x/4)+8..........8.................5.....................2...etc..

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Let S be the basis {1,t,t^2} of P2.BETTER CALL THIS Q2 TO AVOID CONFUSION WITH P2 BELOW
Set
P1= 2+t^2
P2= 1-t+t^2
P3= 3-t+t^2
in P2.
Show that P1,P2,P3 isa basis of Q2 and find the coordinates of 1,t,t^2 in this new basis.
TO SHOW THAT P1,P2,P3 IS A BASIS,WE NEED TO PROVE THAT THEY ARE INDEPENDENT.
THAT IS IF
AP1+BP2+CP3=0...THEN A=B=C=0
AP1+BP2+CP3=A(2+T^2)+B(1-T+T^2)+C(3-T+T^2)=0
T^2(A+B+C)-T(B+C)+(2A+B+3C)=0.....................I..........HENCE
A+B+C=0 ............................II
B+C=0...............................III
2A+B+3C=0...........................IV
EQN.II - EQN.III..GIVES.....A=0......V...SUBSTITUTING IN IV,WE GET
B+3C=0...............................VI
EQN.VI-EQN.III...GIVES....2C=0...OR...C=0..WHICH ON SUBSTITUTING IN EQN.I GIVES
B==0...SO A=B=C=0...HENCE P1,P2,P3 ARE INDEPENDENT.HENCE THEY FORM A BASIS.
NOW TO WRITE Q2 IN THE EW BASIS WE EQUATE EQN.I TO Q2 AND FIND A,B,C.
T^2(A+B+C)-T(B+C)+(2A+B+3C)=Q2=1+T+T^2...HENCE
A+B+C=1.....................VII
B+C=-1......................VIII
2A+B+3C=1...................IX
PROCEEDING AS BEFORE WE GET
A=2,B=0,C=-1..HENCE
Q2=1+T+T^2=2(2+T^2)-(3-T+T^2)

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find a third-degree polynomial equation with rational coefficients that has the roots -5 and 4+i
IF THE COEFFICIENTS OF POLYNOMIAL ARE REAL ,THEN COMPLEX ROOTS CAN OCCUR ONLY IN CONJUGATES.SO IF 4+I IS A ROOT THEN 4-I IS ALSO A ROOT.SO,WE HAVE ALL THE 3 ROOTS FOR THE 3 RD. DEGREE POLYNOMIAL...HENCE THE POLYNOMIAL IS GIVEN BY...
(X+5)(X-(4+I))((X-(4-I))=0
(X+5){(X-4)^2-I^2}=0
(X+5)(X^2+16-8X+1)=0
=(X+5)(X^2-8X+17)=0
=X^3-8X^2+17X+5X^2-40X+85=0
=X^3-3X^2-23X+85=0

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For what integral values of n is the expression 1^n + 2^n + 3^n + 4^n divisible by 5?
WE CAN EASILY SEE THAT FOR N=1,2,3 ETC..IT IS DIVISIBLE BY 5..FOR EX...FOR N=1,WE HAVE
1^1+2^1+3^1+4^1=1+2+3+4=10 WHICH IS DIVISIBLE BY 5..SIMILARLY FOR N=2,3 ETC..
IF YOU WANT A PROOF OF THE THEOREM USE INDUCTION METHOD ...

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A bag contains 6 red marbles, 9 blue marbles, and 5 green marbles. You withdrew one marble,replace it, and then withdraw another marble. what is the probability that you do NOT pick two green marbles?
in such problems it is easier to find probability of an event happening and subtract it from total probability of 1 to get probability of event not happening.
SO LET US CONSIDER PROBABILITY OF DRAWING 2 GREEN MARBLES.
THERE ARE 6 RED+9 BLUE+5 GREEN MARBLES=20 MARBLES.
PROBABILITY OF GETTING A GREEN MARBLE ON FIRST ATTEMPT=5/20=1/4
PROBABILITY OF GETTING A GREEN MARBLE ON SECOND ATTEMPT=
=PROBABILITY OF GETTING A GREEN MARBLE ON FIRST ATTEMPT AS THE MARBLE IS REPLACED =5/20=1/4
HENCE PROBABILITY OF DRAWING 2 GREEN MARBLES = (1/4)8(1/4)=1/16
HENCE PROBABILITY OF NOT DRAWING 2 GREEN MARBLES =1-1/16=15/16

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ORTHOGONAL MATRICES ARE SQUARE MATRICES SUCH THAT PRODUCT OF AN ORTHOGONAL MATRIX WITH ITS TRANSPOSE IS AN IDENTITY MATRIX.THAT IS IF
A*A'=A'*A=I,THEN A IS AN ORTHOGONAL MATRIX.
TO PROVE THAT THEY FORM A SUB GROUP UNDER THE GENERAL LINEAR GROUP OF MATRICES,WE NEED TO SHOW THAT
I.ORTHOGONAL MATRICES ARE A SUB SET OF THE GENERAL GROUP OF MATRICES.
II.UNDER THE SAME COMPOSITION AS IN THE GENERAL GROUP,THE SUB GROUP ITSELF IS A GROUP.HERE WE CONSIDER THE COMPOSITION OF MULTIPLICATION.THIS MEANS THAT WE HAVE TO SHOW WITHIN THE SUBGROUP FOR MULTIPLICATION......
1.CLOSOURE
2.ASSOCIATIVITY
3.EXISTENCE OF IDENTITY
4.EXISTENCE OF INVERSE
NOW I CRITERIA IS OBVIOUS AS THE SET OF ORTHOGONAL MATRICES (CALLED S) AS DEFINED ARE INDEED A SUB SET OF GENERAL LINEAR GROUP OF MATRICES(CALLED G).
II....MULTIPLICATION IS INDEED WELL DEFINED AND EXISTS FOR ORTHOGONAL MATRICES AS IT IS FOR THE GENERAL GROUP OF MATRICES.
1 AND 2.ASSOCIATIVITY HOLDS FOR MATRIX MULTIPLCATION IN GENERAL SO WE SHALL PROVE CLOSOURE HERE.
LET A AND B BE ORTHOGONAL MATRICES IN THE SUB SET OF ORTHOGONAL MATRICES S .SINCE A' AND B' ARE ALSO ORTHOGONAL MATRICES,THEY ARE ALSO ELEMENTS OF S.
SO A*A'=A'*A=I...AND...B*B'=B'*B=I...
NOW..
IF WE SHOW THAT A*B IS ALSO AN ORTHOGONAL MATIX THEN IT WILL BE AN ELEMENT OF S WHICH PROVES CLOSOURE......
TST (A*B)*(A*B)'=I
WE KNOW FOR MATRIX MULTIPLICATION THAT (A*B)'=(B'*A')...HENCE
(A*B)*(A*B)'=(A*B)*(B'*A')=A*(B*B')*A'=A*I*A'=A*A'=AA'=I
HENCE A*B IS ALSO AN ORTHOGONAL MATRIX.HENCE IT IS AN ELEMENT OF S.
3. IDENTITY ELEMENT IN S IS THE UNIT MATRIX WHICH IS SAME AS IN G.SINCE,WE HAVE
A*I=I*A=A...WHICH FOLLOWS OBVIOUSLY FROM MULTIPLICATIVE PROPERTY.
4.WE HAVE A*A'=A'*A=I.....IF A IS AN ORTHOGONAL MATRIX ITS INVERSE A' IS ALSO AN ORTHOGONAL MATRIX. HENCE IT IS AN ELEMENT OF S.
THUS S IS A SUB GROUP OF G.

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DRAW A CIRCLE.INSCRIBE A TRIANGLE ABC IN IT.DRAW AD BISECTING ANGLE BAC AND MEETING THE CIRCLE AT D.SO IN THE 2 SEGMENTS OF THE CIRCLE..NAMELY ARC BD AND ARC DC WE HAVE,BOTH ARCS SUBTENDING EQUAL ANGLES AT A .SINCE ANGLE BAD=ANGLE DAC.
SO THE ARCS SHALL BE EQUAL IN LENGTH.SO ARC BD = ARC DC
SO D IS THE MID POINT OF ARC BDC

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Find the foci of the ellipse whose equation is (y^2/16)+(x^2/12)=1.
> COMPARING WITH STANDARD EQN. OF ELLIPSE
> (X^2/A^2)+(Y^2/B^2)+1...WE GET
> FOCI=(AE,0) AND (-AE,0)..WHERE E=SQRT.{(A^2-B^2)/A^2}
> WE HAVE A=4 AND E=SQRT.{(16-12)/16}=1/2..SO FOCI ARE..(2,0),(-2,0)

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THERE ARE 4 ACES PLUS 12 FACE CARDS =16 IN ALL IN A PACK OF 52 CARDS
SO PROBABILITY OF DRAWING ONE OF THEM IS =16/52=4/13

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THIS IS GIVEN BY 7P7=7!=7*6*5*4*3*2*1=5040

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Solve the equation 1nx=2. Round the answer to four decimal places.
> LN X=2
> X=E^2=7.389056..FROM CALCULATOR..

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LET 5^X=Y
> 25^X=(5^2)^X=5^2X=(5^X)^2=Y^2
> 125^X=(5^3)^X=5^3X=(5^X)^3=Y^3
WE HAVE....25^(X+1)=125^X...
(25^X)(25)=125^X...OR...
25Y^2=Y^3
Y^2(Y-25)=0...OR...Y=25..
5^X=25...SO..X=2

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Find the midpoint and the distance between the points (6,-8) and (14,7).
DISTANCE BETWEEN (X1,Y1) AND (X2,Y2) IS GIVEN BY
SQRT.{(X2-X1)^2+(Y2-Y1)^2}
AND THEIR MID POINT IS GIVEN BY {(X1+X2)/2,(Y1+Y2)/2}
SO WE GET DISTANCE =SQRT.{(14-6)^2+(7+8)^2}=SQRT(64+225)=SQRT.(289)
MIDPOINT IS {(6+14)/2,(-8+7)/2}=(10,-1/2)

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THERE ARE 26 RED CARDS OUT OF 52.
HENCE PROBABILITY OF DRAWING RED CARD FIRST=26/52=1/2
NOW THERE ARE 26 BLACK CARDS AND 51 TOTAL CARDS
PROBABILITY OF DRAWING BLACK CARD IS =26/51
SO PROBABILITY OF DRAWING FIRST RED AND THEN BLACK CARD =(1/2)(26/51)
THIS ENDS YOUR PROBLEM.
*****************************************************************************
BUT IF PROBILITY OF ONE RED AND ONE BLACK CARD IS NEEDED THEN.....
SIMILARLY PROBABILITY OF DRAWING FIRST BLACK AND THEN RED CARD =(1/2)(26/51)
HENCE TOTAL PROBABILITY OF DRAWING ONE BLACK AND ONE RED CARD =SUM OF THE ABOVE 2 PROBABILITIES.=26/51

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Let a not equal 0, b and c be integers with a and b relatively prime.
Show that if a|b*c
A|BC...HENCE....
BC=A*M......I.......WHERE M IS AN INTEGER
AND B ARE RELATIVELY PRIME...THAT IS THEY HAVE NO COMMON FACTORS OR THEIR GCD IS 1.
HENCE GCD =1 =AX+BY...WHERE X AND Y ARE INTEGERS..MULTIPLYING BY C THROUGHOUT,WE GET..
C=CAX+CBY...SUBSTITUTING EQN.I...
C=CAX+AMY=A(CX+MY)=A*K SAY.....WHERE K=CX+MY...
BUT C,X,M,Y ARE ALL INTEGERS...HENCE K IS AN INTEGER.
C=A*K....OR...A DIVIDES C....A|C

then a|c.
how do i give equations for relatively prime and divisible and then substitute.
USE GCD OF A AND B =XA+YB...TO CONVERT INTO EQNS.HOPE YOU UNDERSTOOD.

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(a,b)*(c,d) = (ac,ad + b)
TO SHOW THIS OPERATION * IS A GROUP,WE NEED TO PROVE...
1..CLOSOURE...
AC IS ALSO REAL NUMBER SINCE A AND C ARE REAL...AD IS ALSO REAL AND SO IS AD+B.SO (AC,AD+B)IS AN ELEMENT OF R^2
2..ASSOCIATIVE....
TPT...{(A,B)*(C,D)}*(E,F)=(A,B)*{(C,D)}*(E,F)}..
LHS=(AC,AD+B)(E,F)=(ACE,ACF+AD+B)
RHS=(A,B)(CE,CF+D)=(ACE,ACF+AD+B)=LHS
3...EXISTENCE OF IDENTITY....LET IT BE (I1,I2)
WE SHOULD HAVE (A,B)*(I1,I2)=(A,B) AND (I1,I2)*(A,B)=(A,B)
AI1=A...SO...I1=1
AI2+B=B...SO....I2=0....SO...(1,0) IS THE IDENTITY ELEMENT..IT IS AN ELEMENT OF R^2...OK....
LET US CHECK...
(1,0)*(A,B)=(1A,1B+0)=(A,B)...OK...
4...EXISTENCE OF INVERSE....
WE SHOULD FIND (X,Y)SO THAT (X,Y)(A,B)=(1,0)=(A,B)*(X,Y)
XA=1....OR..X=1/A.....
XB+Y=0.....Y=-XB=-B/A.....SO INVERSE IS (1/A,-B/A)..SINCE...A IS NOT ZERO ,WE HAVE THE INVERSE AS AN ELEMENT OF R^2.
LET US CHECK...
(A,B)(1/A,-B/A)=(A*1/A,((A*-B)/A)+B)=(1,0) ....OK....
HENCE * AS A BINARY OPERATION IS A GROUP.
WE FIND THAT
(a,b)*(c,d) = (ac,ad + b)..WHERE AS
(C,D)*(A,B) = (CA,CB+D) WHICH ARE NOT EQUAL.
HENCE THIS IS NOT ABELIAN

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For what value of 'c' is 'ax - by= c' a direct variation?
1 solutions
--------------------------------------------------------------------------------
Answer 14515 by venugopalramana(833) on 2006-02-07 00:05:37 (Show Source):
WE SAY X VARIES DIRECTLY AS Y IF X=K*Y,WHERE K IS A CONSTANT..HERE WE HAVE
AX-BY=C..OR...AX=BY+C...OR...X=(BY+C)/A = (B/A)Y+(C/A)..BUT FOR DIRECT PROPORTION X=K*Y...HENCE C/A=0...OR...C=0

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Let r=12*95*1750...WRITE AS PRODUCT OF PRIMES...
R=(2*2*3)*(5*19)*(2*5*5*5*7)=(2^3)(3)(5^4)(7)(19)
and h=38*25^(2)*16^(3).
H=(2*19)((5*5)^2)((2*2*2*2)^3)=(2^13)(5^4)
HENCE GCD =(2^3)(5^4)
LCM=(2^13)(3)(5^4)(7)(19)
COMMON DIVISORS ARE ALL FACTORS OBTAINED FROM
GCD =(2^3)(5^4)...THAT IS (2^0)(5),(2^0)(5^2),(2^0)(5^3),(2^0)(5^4)
(2^1)(5^0),(2^1)(5^1).....ETC........
Give all common divisors, gcd, and lcm of a and b

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LET US COMPARE THE GIVEN SERIES WITH A GEOMETRIC SERIES WHOSE WE CAN DO...
THE G.P IS ...1,2,4,8,16,.......WITH C.R. OF 2 AND A=1.
FIRSTLY WE FIND THAT THE GIVEN SERIES IS ALL POSITIVE NUMBERS AND INCREASING SERIES AFTER F2,SINCE EVERY TERM IS THE SUM OF PREVIOUS 2 TERMS WHICH ARE POSITIVE...JENCE..F2 NOW WE HAVE..
F1=1............................................G1=1.........F1=G1
F2=1............................................G2=2G1.......F2 F3=F1+F2 F4=F2+F3 ....................................................................
.....................................................................
FN=F(N-2)+F(N-1) ------------------------------------------------------------------------
ADDING ALL THE ABOVE WE GET...(F1+F2+F3+F4+...+FN)<(G1+G2+G3+.....+GN)
BUT WE KNOW THAT
G1+G2+G3+.....+GN=1+2+4+8+.........2^(N-1)
A=1...R=2...SO...SUM IS ..A*(R^N-1)/(R-1)=1*(2^N-1)/(2-1)=2^N-1
HENCE
(F1+F2+F3+F4+...+FN)<(G1+G2+G3+.....+GN)=2^N-1<2^N...THEN OBVIOUSLY
FN<2^N

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Its from differential calculus.
LEt y= 5/³√(2x-5)^2
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)

Your Answer:
see my working and comments on your work-----------
--------------------------------------------------------
Its from differential calculus.
LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N)
Y=5*(2X-5)^(-2/3)...
DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1)
DY/DX=(-20/3)*(2X-5)^(-5/3)
(-20)/3*(2X-5)^(5/3)
------------------------------------------------------
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT +
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)

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WE SAY X VARIES DIRECTLY AS Y IF X=K*Y,WHERE K IS A CONSTANT..HERE WE HAVE
AX-BY=C..OR...AX=BY+C...OR...X=(BY+C)/A = (B/A)Y+(C/A)..BUT FOR DIRECT PROPORTION X=K*Y...HENCE C/A=0...OR...C=0

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Its from differential calculus.
LEt y= 5/³√(2x-5)^2
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)

Your Answer:
see my working and comments on your work-----------
--------------------------------------------------------
Its from differential calculus.
LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N)
Y=5*(2X-5)^(-2/3)...
DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1)
DY/DX=(-20/3)*(2X-5)^(-5/3)
(-20)/3*(2X-5)^(5/3)
------------------------------------------------------
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT +
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)

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see my working and comments on your work-----------
--------------------------------------------------------
Its from differential calculus.
LEt y= 5/³√(2x-5)^2 ..USE X^(-N)=1/(X^N)AND N th. ROOT OF X =X^(1/N)
Y=5*(2X-5)^(-2/3)...
DY/DX=5*(2X-5)^(-2/3-1)*2..USING DY/DX=DY/DU*DU/DX..HERE U=2X-5 AND DERIVATIVE OF X^N = N*X^(N-1)
DY/DX=(-20/3)*(2X-5)^(-5/3)
(-20)/3*(2X-5)^(5/3)
------------------------------------------------------
So far I got upto:
d/dx=5(2x^2-10x+25)^(2/3)..YOU SQUARED 2X-5 ..THEN WHY AGAIN POWER OF 2/3..IT SHOULD BE ONLY 1/3..FURTHER IT SHOULD BE -1/3 AND NOT +
However, I can not continue on from there.
the answer at the back on my book is:
-20
----------
3(2x-5)^(5/3)
You may edit the question. Maybe convert formulae to the same formula notation as in your solutions.

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log^9(3x+14)-log^95=log^92x.
i think you mean all logs are to base 9...in that case since all are to a common base we can drop the base.
log(3x+1)-log 5=log(2x)....or.....log(3x+1)-log 5-log(2x)=0
log{(3x+1)/(5*2x)=0..taking antilogs,we get
(3x+1)/10x=1
3x+1=10x...or...10x-3x=1...
7x=1
x=1/7