সূচক ও লগারিদম (303 টি প্রশ্ন )
log₁₀(1000) = x
⇒ log₁₀10³ = x
⇒ 3log₁₀10 = x
⇒ x = 3
log2 + log4 + log16 + log256 + ...
প্রথম 10টি পদ হবে,
log(2¹) + log(2²) + log(2⁴) + log(2⁸) + log(2¹⁶) + log(2³²) + log(2⁶⁴) + log(2¹²⁸) + log(2²⁵⁶) + log(2⁵¹²)
= = (1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512)log2
= (2¹⁰ - 1)log2
(19)12 × (19)8 ÷ (19)4 
= (19)12 × (19)8 / (19)
= (1912 + 8) / (19)
= 1920/194
= 1920 - 4
= 1916

∴ (19)12 × (19)8 ÷ (19)4 = (19)16
মনে করি, 
log36 X = A
⇒ 36A = X
⇒ (62)A = X
⇒ 62A = X
⇒ log6 X = 2A
√√√√x = xm/n
⇒ √√√((x)1/2)= xm/n
⇒ (((((x)1/2)1/2)1/2)1/2) = xm/n 
⇒ (x)1/16= xm/n 
⇒ m/n = 1/16
⇒ m = n/16
log381
= log334
= 4log33
= 4 × 1
= 4
log273√3
= log2733/2
= log2733.(1/2)
= log27271/2
= 1/2 log2727
= 1/2
2log105 + log108 - (1/2)log104
=log1052 + log108 - (1/2)log10(22)
=log1025 + log108 - log10(22)1/2
=log1025 + log108 - log102
=log10{(25 × 8)/2}
=log10100
=log10102
=2log1010
=2 . 1
= 2
(81)0.16 × (81)0.09
= (81)(0.16 + 0.09)
= (81)0.25
=(81)1/4
= (34)(1/4)
= 3
 4(x - y) = 64
বা,  4(x - y) = 43
বা, x - y = 3 ........................(1)

আবার,
4(x + y) = 45
বা, x + y = 5 .......................(2)

(1) + (2) হতে পাই,
2x = 8
∴ x = 4
log₅(125x) - log₅(25) = 1
log₅(125x/25) = 1
log₅(5x) = 1
5x = 5¹
x = 1
log10(x + 3) = log10x + log103
বা, log10(x + 3) = log10(x × 3)
বা, log10(x + 3) = log103x
বা, x + 3 = 3x
বা, 2x = 3
∴ x = 3/2
log2 8 + 3/2 log3 3√9 
= log2 23 +3/2 log3 (32)1/3
= 3 log2 2 + 3/2 log3 32/3
= 3 × 1 + 3/2 . 2/3  log3 3
= 3 + 1 × 1
= 4
22. 32n + 2 - 9n + 1
= 4 . 32(n + 1) - 32(n + 1)
= 32(n + 1) (4 - 1)
= 32n + 2 . 3
= 32n + 2 + 1
= 32n + 3

xx0 + yy0 
= x1 + y1 [অর্থাৎ, a0 = 1, যেখানে a ≠ 0]
= x + y

(5n+2+35×5n-1)/4×5n
= {5n.52+35×5n×(1/5)}/4×5n
= 5n(25+7)/4×5n
= 32/4
= 8


log 6 = log(2 × 3)
= log 2 + log 3
= 0.3010 + 0.4771
= 0.7781
log216
=log224
=4log22
=4.1
=4
log27
= log273√27
= log27271/3
= 1/3 log27
[ ∴ logaMr = rlogaM]
= 1/3 × 1
[ ∴logaa = 1]
= 1/3
(9x - 4)/(3x - 2) -2
= {(3x)2 - (2)2}/(3x - 2) -2
= {(3x + 2) × (3x - 2)}/(3x - 2) -2
= 3x + 2 - 2
= 3x
(3/2)x = 1
⇒ (3/2)x = (3/2)0
[∴ a0 = 1, a > 0]
∴ x = 0
x√0.09 = 3
⇒ x = 3/√0.09
⇒ x2 = 32/0.09
[বর্গ করে]
⇒ x= 9 × 100/9
⇒ x2 = 102
∴ x = 10
2log105 + log1036 - log10
= log1052 + log1062- log1032 
   [ যেহেতু, plogkM = logkMp]
= log10 {(52×62)/32}
   [যেহেতু, logkM + logkN = logk(MN), logkM – logkN = logk(M/N)]
= log10{(25 × 36)/9}
= log10100
= log10102
= 2log1010
= 2.1 [logaa=1]
= 2

logx(1/27) = - 2 
বা, x - 2 = 1/27
বা, 1/x2 = 1/27
বা, x2 = 27
বা, x = √27
বা, x = √(9 × 3)
বা, x = 3√3
বা, x/3 = 3√3/3
∴ x/3 = √3
xz = y3
 ⇒ (101.4)z = (100.7)3
⇒ 101.4z = 102.1
⇒ 1.4z = 2.1
⇒ z = 2.1/1.4
⇒ z = 21/14
∴ z = 1.5
দেয়া আছে,  
x = ya, y = zb এবং z= xc

এখন, 
z = xc
⇒ xc = z
⇒ (ya)c = z
⇒ yac = z
⇒ (zb)ac = z
⇒ zabc = z
⇒ zabc = z1
⇒ abc = 1

(xp/xq)p + q . (xq/xr)q + r. (xr/xp)r + p 
= (xp - q)p + q. (xq - r)q + r . (xr - p)r + p
= x(p - q)(p + q). x(q - r)(q + r) . x(r - p)(r + p)
=xp2 - q2 . xq2 - r2.xr2 - p2 
=xp2 - q+ q2 - r2+r2 - p2
= x0
= 1


log10(0.001)

=  log10(10)-3
= -3×og10(10)
= (-3)*1
= -3

 
9x + 3 = 27x + 1

⇒ (32)x + 3 = (33)x + 1

⇒ 32x + 6 = 33x + 3

⇒ 2x + 6 = 3x + 3

⇒ 3x - 2x = 6 - 3

⇒ x = 3
A ÷ B = A/B 
=(81)x - 1/9x - 1
= (92)x - 1/9x - 1
= { (9x + 1) (9x - 1)}/(9x - 1)
= 9x + 1
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