O Level A-Maths Logarithm Problems

Mugen Benoit4real SG

Hi, I have got series of logarithm questions that I could not solve:

Q1) Given that lg 2 = a and lg3 = b, express lg cube root 972 in terms of a and b. Find x if lg x = 3a - 4b + 1.
A1) 2a + 5b / 3, 80/81

Q2) 2 log2 X = log3 81 + log2 (X + 5)
A2) 20

Attempt Q2)
2 log2 X = Log3 81 + log2 (X + 5)
log2 X^2 = log3 81 + log2 (X + 5)
log X^2/log 2 = log 81/log 3 + log (X + 5)/log 2
log X^2/log 2 = 4 + log (X + 5)/log 2
log X^2/log 2 - log (X + 5)/log 2 = 4 <- from this point, I do not know how to solve

Q3) 4e^2y - 21 = 0
A3) 0.829

Attempt Q3)
4e^2y - 21 = 0
4e^2y = 21
2y ln 4e = ln 21 <- from this point, I do not know how to solve

Q4) 1/logx 2 + 1/logx 3 + 1/logx 6 = 3.6
A4) 3.44

Attempt Q4)
1/logx 2 + 1/logx 3 + 1/logx 6 = 3.6
log2 X + log3 X + log6 X = 3.6
log X/log 2 + log X/log 3 + log X/log 6 = 3.6 <- from this point, I do not know how to solve

Q5) Solve the simultaneous equation e square root e^x = e^2y and log4 (X + 2) - log2 Y = 1
A5) x=2, y=1

Attempt Q5)
From: e square root of e^x = e^2y
(e square root of e^x)^2 = (e^2y)^2
e^2 . e^x = e^e^9y
e^x = e^4y/e^2
X - 4y = -2

From log4 (X + 2) = log2 y = 1
log (X + 2)/log 4 - log y/log 2 = 1 <- from this point, I do not know how to solve

wee_ws SG

Hi,

For Q2, it is easier to proceed by combining expressions with the same base:

log_2 { x^2 / (x + 5) } = log_3 (3^4)

=> x^2 / (x + 5) = 4^2,

from which we solve a factorisable quadratic equation, noting that x > 0.

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

For Q3:

e^(2y) = 21/4

=> e^y = sqrt(21/4), since e^y > 0 for any real y

=> y = ln { sqrt(21/4) }, computable with a calculator.

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

For Q4, we apply change of base (to e) for each term on the LHS:

log_x 2 = (ln 2) / (ln x),

log_x 3 = (ln 3) / (ln x),

log_x 6 = (ln 6) / (ln x),

so that the original equation becomes

(ln x) / (ln 2) + (ln x) / (ln 3) + (ln x) / (ln 6) = 3.6

=> (ln x) { 1 / (ln 2) + 1 / (ln 3) + 1 / (ln 6) } = 3.6,

from which x could be found with help from the calculator.

Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

For Q5:

sqrt(e^x) = e^(2y)

=> e^x = e^(4y)

=> x = 4y.

However, your answer (that x = 2, y = 1) does not satisfy the equation. Perhaps you may like to check further. Thanks.

Cheers,
Wen Shih

wee_ws SG

Hi,

For Q1:

1. Note that 972 = (2^2)(3^5).

2. 3a - 4b + 1 = 3 lg 2 - 4 lg 3 + lg 10
= lg (2^3) - lg (3^4) + lg 10
= lg { (8/81)(10) }.

Thanks.

Cheers,
Wen Shih

Mugen Benoit4real SG

Hi,

For Q5:

sqrt(e^x) = e^(2y)

=> e^x = e^(4y)

=> x = 4y.

However, your answer (that x = 2, y = 1) does not satisfy the equation. Perhaps you may like to check further. Thanks.

Cheers,
Wen Shih

For this part, it is e sqrt(e^x) = e^(2y) in e in front of the square root is mssing which I have done:

From: e sqrt(e^x) = e^2y
(e sqrt(e^x))^2 = (e^2y)^2
e^2 . e^x = e^e^9y
e^x = e^4y/e^2
X - 4y = -2

I have doubt in this equation which is from Q5:

From log4 (X + 2) = log2 y = 1
log (X + 2)/log 4 - log y/log 2 = 1 <- from this point, I do not know how to solve

Thanks for your help as always ;)

wee_ws SG

Hi,

Sorry, me cock-eyed :P

Given: e . sqrt(e^x) = e^(2y)

=> e^2 . e^x = e^(4y)

=> 2 + x = 4y

=> x - 4y = -2 -- (1).

Given: log_4 (x + 2) - log_2 y = 1

=> log_4 (x + 2) - { (log_4 y) / (log_4 2) } = 1, by a change of base in the 2nd term

=> log_4 (x + 2) - 2 log_4 y = 1, since log_4 2 = log_4 (4^(1/2))

=> log_4 { (x + 2) / (y^2) } = 1, since log a - log b = log (a/b) and k log a = log (a^k)

=> (x + 2) / (y^2) = 4 -- (2).

You should be able to proceed from this point, jiayou :)

Cheers,
Wen Shih

Mugen Benoit4real SG

Thanks for your great help! :D