Operation of Indices

We have :

\(=\dfrac{x+2}{89}+1+\dfrac{x+5}{86}+1>\dfrac{x+8}{83}+1+\dfrac{x+11}{80}+1\)

\(=\dfrac{x+91}{89}+\dfrac{x+91}{86}>\dfrac{x+91}{83}+\dfrac{x+91}{80}\)

\(=\dfrac{x+91}{89}+\dfrac{x+91}{86}-\dfrac{x+91}{83}-\dfrac{x+91}{80}>0\)

\(=\left(x+91\right)\left(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\right)>0\)

To \(\dfrac{1}{89}+\dfrac{1}{86}-\dfrac{1}{83}-\dfrac{1}{80}\) different 0

\(=x+91>0\)

\(=x< -91\)

e have :

=x+289+1+x+586+1>x+883+1+x+1180+1

=x+9189+x+9186>x+9183+x+9180

=x+9189+x+9186−x+9183−x+9180>0

=(x+91)(189+186−183−180)>0

To 189+186−183−180

different 0

=x+91>0

=x<−91

a + b = 5

<=> (a+b)3=125

 <=> a3 + 3a2b + 3ab2 + b3 = 125

<=> 3ab(a + b) + 35 = 125 (a3 + b3 = 35)

<=> 15ab = 90 (a + b = 5)

<=> ab = 6

a + b = 5

<=> (a + b)2 = 25

<=> a2 + 2ab + b2 = 25

<=> a2 + 12 + b2 = 25

<=> a2 + b2 = 13

a + b = 5

<=> \(\left(a+b\right)^3=125\) <=> a³ + 3a²b + 3ab² + b³ = 125

<=> 3ab(a + b) + 35 = 125 (a³ + b³ = 35)

<=> 15ab = 90 (a + b = 5)

<=> ab = 6

a + b = 5

<=> (a + b)² = 25

<=> a² + 2ab + b² = 25

<=> a² + 12 + b² = 25

<=> a² + b² = 13

We have : (x + y)2 - (x - y)2 = 18 - 12 

(x2 + 2xy + y2) - (x2 - 2xy + y2) = 6

                                       4xy       = 6

                                       2xy       = 3

=> x2 + y2 = 18 - 3 = 15

We have : (x + y)² - (x - y)² = 18 - 12 

(x² + 2xy + y²) - (x² - 2xy + y²) = 6

                                       4xy       = 6

                                       2xy       = 3

=> x² + y² = 18 - 3 = 15

33x+2y=33x×32y=(3x)3×(3y)2=(29)3×272

=23×3636=8

\(3^{3x+2y}=3^{3x}\times3^{2y}=\left(3^x\right)^3\times\left(3^y\right)^2=\left(\dfrac{2}{9}\right)^3\times27^2\)

\(=\dfrac{2^3\times3^6}{3^6}=8\)

We have : (2x2 + 1)(5x + 2) = 10x3 + 4x2 + 5x + 2

So a = 4

We have : (2x² + 1)(5x + 2) = 10x³ + 4x² + 5x + 2

So a = 4

(a) 5m.252m = 1252m - 1

=> 5m.54m = (53)2m - 1

=> 55m = 56m - 3

=> 5m = 6m - 3

=> m = 3

(b) 5m.25m.125m.625m = 530

=> (5.25.125.625)m = 530

=> (5.52.53.54)m = 530

=> 510m = 530

=> 10m = 30

=> m = 3

(a) 5^m.25^2m = 125^{2m - 1}

=> 5^m.5^4m = (5³)^{2m - 1}

=> 5^5m = 5^{6m - 3}

=> 5m = 6m - 3

=> m = 3

(b) 5^m.25^m.125^m.625^m = 5³⁰

=> (5.25.125.625)^m = 5³⁰

=> (5.5².5³.5⁴)^m = 5³⁰

=> 5^10m = 5³⁰

=> 10m = 30

=> m = 3

Because 5x2−4x=5x2−4x and 2>−3

 => a > b 

You are right (True).

Because \(5x^2-4x=5x^2-4x\) and \(2>-3\) => a > b 

You are right (True).

(x + y)2 = 9

<=> x2 + 2xy + y2 = 9 (1)

(x - y)2 = 1

<=> x2 - 2xy + y2 = 1 (2)

(1) - (2) => 4xy = 8 => xy = 2

=> x2 + 2xy + y2 = 9

<=> x2 + 4 + y2 = 9

<=> x2 + y2 = 5

(x + y)² = 9

<=> x² + 2xy + y² = 9 (1)

(x - y)² = 1

<=> x² - 2xy + y² = 1 (2)

(1) - (2) => 4xy = 8 => xy = 2

=> x² + 2xy + y² = 9

<=> x² + 4 + y² = 9

<=> x² + y² = 5

We have : 725 = (74)6.7

The units digit of 74 is 1 (74 = 2401),so the units digit of (74)6 is 1

=> The units digit of 725 is 7

We have : 7²⁵ = (7⁴)⁶.7

The units digit of 7⁴ is 1 (7⁴ = 2401),so the units digit of (7⁴)⁶ is 1

=> The units digit of 7²⁵ is 7

By the assumption  \(\left\{{}\begin{matrix}a+b=4\\a^3+b^3=28\end{matrix}\right.\Rightarrow a^2-ab+b^2=\dfrac{28}{4}=7\Rightarrow2a^2-2ab+2b^2=14\).

Other hand, \(a+b=4\Rightarrow a^2+2ab+b^2=16\). So:

           \(3\left(a^2+b^2\right)=\left(2a^2-2ab+2b^2\right)+\left(a^2+2ab+b^2\right)=14+16\Rightarrow a^2+b^2=10\)

By the assumption  {a+b=4a3+b3=28⇒a2−ab+b2=284=7⇒2a2−2ab+2b2=14

.

Other hand, a+b=4⇒a2+2ab+b2=16

. So:

           3(a2+b2)=(2a2−2ab+2b2)+(a2+2ab+b2)=14+16⇒a2+b2=10

\(m^2+n^2=\left(m+n\right)^2-2mn=103\)

m2+n2

= (m+n)2-2mn

= 112-2.9

=121-18

=103

m²+n²

= (m+n)²-2mn

= 11²-2.9

=121-18

=103

\(27^x.243^y=\left(3^3\right)^x\left(3^5\right)^y=3^{\left(3x+5y\right)}=3^2=9\)

27x.243y=(33)x(35)y=3(3x+5y)=32=9

We have    \(a^m=\dfrac{1}{3},a^n=6\Rightarrow2=6.\dfrac{1}{3}=a^n.a^m=a^{m+n}\), so  

                                                     \(2^{2m+3n}=\left(a^{m+n}\right)^{2m+3n}=a^{\left(m+n\right)\left(2m+3n\right)}\) .

We have    am=13,an=6⇒2=6.13=an.am=am+n

, so  

                                                     22m+3n=(am+n)2m+3n=a(m+n)(2m+3n)

 .

Put   \(a=2^{777},b=3^{555},c=4^{444}\). We have :

\(a=2^{777}=\left(2^7\right)^{111}=128^{111},b=3^{555}=\left(3^5\right)^{111}=243^{111},c=4^{444}=\left(4^4\right)^{111}=256^{111}\Rightarrow a< b< c\).

So:  \(2^{777}< 3^{555}< 4^{444}\).

Put   a=2777,b=3555,c=4444

. We have :

a=2777=(27)111=128111,b=3555=(35)111=243111,c=4444=(44)111=256111⇒a<b<c

.

So:  2777<3555<4444

.

a) \(3^x=243=3^5\Leftrightarrow x=5\).

b) \(5^m=625\Leftrightarrow5^m=5^4\Leftrightarrow m=4\).

c) By  \(16^x.5^{2x}=16^x.25^x=\left(16.25\right)^x=400^x\), so the done equation equivalant to   

                                                          \(400^x=400^1\Leftrightarrow x=1\).

d) \(2^{2n+2}+2^{2n}=160\Leftrightarrow2^{2n}\left(2^2+1\right)=160\Leftrightarrow2^{2n}=2^5\Leftrightarrow2n=5\Leftrightarrow n=2,5.\)

a) 3x=243=35⇔x=5

.

b) 5m=625⇔5m=54⇔m=4

.

c) By  16x.52x=16x.25x=(16.25)x=400x

, so the done equation equivalant to   

                                                          400x=4001⇔x=1

.

d) 22n+2+22n=160⇔22n(22+1)=160⇔22n=25⇔2n=5⇔n=2,5.

b) $5^m=625\iff 5^m=5^4\iff m=4$.

c) By  ${16}^x{.5}^{2x}={16}^x{.25}^x={\left(16.25\right)}^x={400}^x$, so the done equation equivalant to   

                                                          ${400}^x={400}^1\iff x=1$.

d) $2^{2n+2}+2^{2n}=160\iff 2^{2n}\left(2^2+1\right)=160\iff 2^{2n}=2^5\iff 2n=5\iff n=2,5.$

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