Kindly solve q4a&b and q6 c&d. Thanks

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# Answer

**Solution for 6(c)**

log_{x} 8+ 2 log_{2} x = 7

(log_{2} 8) / (log_{2} x ) + 2 log_{2} x = 7

3 / (log_{2} x) + 2 log_{2} x = 7

Multiply both sides by log_{2} x to remove it from the denominator on LHS

3 + 2 (log_{2} x)^{2} = 7log_{2} x

Let A = log_{2} x

2A^{2} – 7A + 3 = 0

(2A – 1) (A – 3) = 0

=> A = 1/2 or A = 3

log_{2} x = 1/2 => x = √ 2 log_{2} x = 3 => x = 8

*****************************************************************

**Check by substituting both x into the original equation:**

log_{x} 8+ 2 log_{2} x = 7

When x = √ 2 ,

LHS = log_{√ 2 } 8 + 2 log_{2} √ 2 = 6 + log_{2} (2)^{(1/2)x2} = 6 + 1 = 7 = RHS

(√ 2)^{(6)} = 8

log_{2} (2) = 1

When x = 8 ,

LHS = log_{8} 8+ 2 log_{2} 8 = 1 + log_{2} (8)^{2} = 1 + 6 = 7 = RHS

(2)^{6} = (2^{3} )^{2 } = 64

that is the part when i multiply both side, i got 3 log2 X and that was wrong it has to be 4 logx X. thanks Sushi88

You are welcome! 🙂 Good that you understand where you have gone wrong. Practice makes perfect, it would become easier and easier for you.

You need to upload better quality pictures.

Can see better now?

how about 6c? thanks , tried 6 times, still cannot get it.

6c question is logx 8 + 2 log2 X =7

**Solution for 6(c)**

log_{x} 8+ 2 log_{2} x = 7

(log_{2} 8) / (log_{2} x ) + 2 log_{2} x = 7

3 / (log_{2} x) + 2 log_{2} x = 7

Multiply both sides by log_{2} x to remove it from the denominator on LHS

3 + 2 (log_{2} x)^{2} = 7log_{2} x

Let A = log_{2} x

2A^{2} – 7A + 3 = 0

(2A – 1) (A – 3) = 0

=> A = 1/2 or A = 3

log_{2} x = 1/2 => x = √ 2 log_{2} x = 3 => x = 8

*****************************************************************

**Check by substituting both x into the original equation:**

log_{x} 8+ 2 log_{2} x = 7

When x = √ 2 ,

LHS = log_{√ 2 } 8 + 2 log_{2} √ 2 = 6 + log_{2} (2)^{(1/2)x2} = 6 + 1 = 7 = RHS

(√ 2)^{(6)} = 8

log_{2} (2) = 1

When x = 8 ,

LHS = log_{8} 8+ 2 log_{2} 8 = 1 + log_{2} (8)^{2} = 1 + 6 = 7 = RHS

(2)^{6} = (2^{3} )^{2 } = 64

that is the part when i multiply both side, i got 3 log2 X and that was wrong it has to be 4 logx X. thanks Sushi88

You are welcome! 🙂 Good that you understand where you have gone wrong. Practice makes perfect, it would become easier and easier for you.

Cannot see clearly, need to load a clearer copy and what’s the answer? Sometimes the answer could be wrong if you have tried 6 times?

Answer for 4a is y=1/16x^4

4b y=(27x/5)^1/2

6d is correct x=1.5 or 6

Can you clarify 4a again? Thanks for your help

I have the same problem as Adwin, we cannot read your questions properly due to the blur image, hence we did not provide the solution.

Here’s my **Solution for 4(b)** blur-print question:

3 log_{8} y = 2 log_{4} x – log_{2} 5y + 3 log_{4} 9

log_{2} y^{3} / log_{2} 8 = (log_{2} x^{2} / log_{2} 4 ) – log_{2} 5y + (log_{2} 9^{3} / log_{2} 4 )

(1/3) log_{2} y^{3 } = (1/2) log_{2} x^{2} – log_{2} 5y + (1/2) log_{2} 9^{3} *<— [ 9 ^{3 } = (3^{2} ) ^{3} ) and log_{b}(x ^{y}) = ylog_{b}(x)]*

log_{2} y = log_{2} x – log_{2} 5y + log_{2} 3^{3}

log_{2} y = log_{2} [ (x)(27) / 5y ]

y^{2} = (27x) / 5

y = ± √ [(27x)/5 ]

Since y cannot be negative, – √ [(27x)/5 ] is extraneous

Hence y = √ [(27x)/5 ]

Can help me on 5d? answer is 2.71 thanks

__Solution for 5(d) __

ln 3x = (2 lg 5)(log_{4} 8)

lg 3x / lg e = (lg 25) (lg 8 / lg 4)

lg 3x = [ (lg 25) (lg 8 / lg 4) ] x lg e = 0.9106

3x = 10^{0.9106}

x = 10^{0.9106 } / 3 = 2.71 ( to 3 s.f)

Yes, the blur picture misled both of us. we saw the log_{9} as log_{3}.

**Solution for 4(a)**

log_{3} 2y – 3 log_{9} y = 2log_{3} 4x – 3log_{27} 2

log_{3} 2y – (log_{3} y^{3} ) / (log_{3} 9) = log_{3} (4x)^{2} – 3 [ (log_{3} 2) / (log_{3} 27 ) ]

log_{3} 2y – (1/2) log_{3} y^{3} = log_{3} 16x^{2} – log_{3} 2

log_{3 } [ 2y / (y)^{3/2}] = log_{3} (16x^{2} /2)

2 / (y) ^{1/2} = 8x^{2}

Square both sides to remove the square root of y,

4 / y = 64x^{4}

y = 1 / 16x^{4}

4(a) is the same as Adwin’s solution

6(d) has a slight variation as we used different rules to get to the same end result.

**Alternative solution for 6(d)**

log_{3} (x + 3) – log_{9} (x/2) = log_{4} 8

log_{3} (x + 3) – log_{3} (x/2) / log_{3} 9 = log_{2} 8 / log_{2} 4

log_{3} (x + 3) – (1/2) log_{3} (x/2) = 3/2

log_{3} [ (x + 3) / (x/2)^{1/2} ] = 3/2

(x + 3) / (x/2)^{1/2 } = 3^{3/2}

Square both sides to get rid of the square root,

(x + 3)^{2} / (x/2) = 27

x^{2} + 6x + 9 = (27/2) x

2x^{2} + 12x + 18 – 27x = 0

2x^{2} – 15x + 18 = 0

(2x – 3)(x – 6 ) = 0

x = 3/2 or x = 6

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