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CAT 2025 Lesson : Logarithm - Additional Problems

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5. Additional Examples

Example 11

Solve for xx where log2log33x\mathrm{log}_{2}\mathrm{log} _{3} {3x} == 0
(11) 11            (22) 44            (33) 99            (44) 1616           

Solution

log2log33x\mathrm{log} _{2} \mathrm{log} _{3} {3x} == 00
⇒ log33x\mathrm{log} _{3} {3x} == 11
⇒ 3x3x == 33 ⇒ xx == 11

Answer: (11) 11


Example 12

If log100.04\mathrm{log} _{10} {0.04} == −1.398-1.398, then log10400\mathrm{log} _{10} {400} == ?
(11)1.6021.602            (22)2.3982.398            (33) 2.6022.602            (44) 3.3983.398           

Solution

log10400\mathrm{log} _{10} {400} == log10(0.04×10000)\mathrm{log} _{10} {(0.04 \times 10000)}
== log100.04+log10104\mathrm{log} _{10} {0.04} + \mathrm{log} _{10} {10^4}
== −1.398+4-1.398 + 4 == 2.6022.602

Answer: (33) 2.6022.602


Example 13

If x=log102 x = \mathrm{log} _{10} {2} and y=log105 y = \mathrm{log} _{10} {5} , then log52\mathrm{log} _{5} {2} == ?
(11) 2y−x2y+x\dfrac{2y - x}{2y +x}           (22)2x+y2x−y\dfrac{2x + y}{2x - y}           (33) 2−x−2y1−x\dfrac{2 - x - 2y}{1 - x}           (44) 2−2x−y1−y\dfrac{2 - 2x - y}{1 - y}          

Solution

By substituting in each of the options we find that only option (3) satisfies

=2−x−2y1−x= \dfrac{2 - x - 2y}{1 - x}

=2−(log 2+2log 5)1−log 2= \dfrac{2 - (\mathrm{log} \ 2 + 2\mathrm{log} \ 5)}{1 - \mathrm{log} \ 2}

=2−log(2×52)1−log 2= \dfrac{2 - \mathrm{log} (2 \times 5^{2})}{1 - \mathrm{log} \ 2} =log 100−log 50log 10−log 2= \dfrac{\mathrm{log} \ 100 - \mathrm{log} \ 50}{\mathrm{log} \ 10 - \mathrm{log} \ 2} =log 2log 5=log52=\dfrac{\mathrm{log} \ 2}{\mathrm{log} \ 5} = \mathrm{log}_{5} {2}

Answer: (33) 2−x−2y1−x\dfrac{2 - x - 2y}{1 - x}


Example 14

If xx == log9144\mathrm{log} _{9} {144} and yy == log123\mathrm{log} _{12} {3}, then xx == ?
(11) 1y\dfrac{1}{y}           (22) 32y\dfrac{3}{2y}           (33) 2y3\dfrac{2y}{3}           (44) 2y2y          

Solution

xx == log32122\mathrm{log} _{3^2} {12^2} == 22log312\dfrac{2}{2} \mathrm{log} _{3} {12} == 1log123\dfrac{1}{\mathrm{log} _{12} {3}} == 1y\dfrac{1}{y}

Answer: (11) 1y\dfrac{1}{y}


Example 15

logx+log(2x+6)\mathrm{log} {x} + \mathrm{log} {(2x + 6)} == log(6−5x)\mathrm{log} {(6 - 5x)}. Solve for xx.
(11) 12\dfrac{1}{2}           (22) −6-6           (33) 12\dfrac{1}{2} or −6-6           (44) None of the above          

Solution

log(2x2+6x)\mathrm{log} {(2x^2 + 6x)} == log(6−5x)\mathrm{log} {(6 - 5x)}
⇒ 2x2+11x−6 2x^2 + 11x - 6 == 00 ⇒ −6-6 or 12\dfrac{1}{2}

Log cannot be applied on a -ve number. At xx == −- 6 6, log xx will be negative and is, hence, rejected.

Answer: (11) 12\dfrac{1}{2}


Example 16

If log2,log(22x+3) \mathrm{log} {2}, \mathrm{log} {(2^{2x} + 3)} and log(22x+27)\mathrm{log} (2^{2x} + 27) are in arithmetic progression, then xx == ?
(11) log23\mathrm{log} _{2} {3}           (22) 2log232 \mathrm{log} _{2} {3}           (33) log25\mathrm{log} _{2} {\sqrt{5}}          (44) 2log252 \mathrm{log} _{2} {5}          

Solution

Let 22x2^{2x} == yy
⇒ 2log(y+3) 2 \mathrm{log} {(y+3)} == log[2×(y+27)]\mathrm{log} {[2 \times (y+27)]}
⇒ log(y2+6y+9)\mathrm{log} {(y^2 + 6y +9 )} == log(2y+54)\mathrm{log} {(2y + 54)}
⇒ y2+4y−45y^2 + 4y - 45 == 00
⇒ yy == 55 or −9-9 [−9-9 is rejected.]

Where x=5x = 5, yy == 22x2^{2x} == 22log222^{2 \mathrm{log} _{2} {\sqrt{2}}} == 2log252^{\mathrm{log} _{2} {5}} == 55

Answer: (33) log25\mathrm{log} _{2} {\sqrt{5}}


Example 17

How many values of xx satisfy logx2−2x+1[logx2+x+1(3x2−6x+7)]=0\mathrm{log}_{x^2 - 2x +1}[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = 0
(1) 0               (2) 1              (3) 2              (4) More than 2

Solution

logx2−2x+1[logx2+x+1(3x2−6x+7)]=0\mathrm{log}_{x^2 - 2x +1}[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = 0

Removing the first logarithm,
⇒ [logx2+x+1(3x2−6x+7)]=(x2−2x+1)0[\mathrm{log}_{x^2+x +1} {(3x^2 - 6x +7)}] = (x^2 - 2x +1)^0

⇒[logx2+x+1(3x2−6x+7)]=1[log_{x^2+x+1} {(3x^2 - 6x +7)}] = 1

Removing the next logarithm,
⇒ (3x2−6x+7)=(x2+x+1)1(3x^2 - 6x +7) = (x^2 + x+1)^1

⇒ 2x2−7x+6=02x^2 - 7x +6 = 0

Solving the quadratic equation, we get
⇒ xx = 32\dfrac{3}{2}, 2

It seems as though 2 solutions exist. However, there are variables in the bases and the value. We need to substitute these solutions and check if log holds good, i.e., logba\mathrm{log}_{b} {a} is defined where aa > 0, bb > 0 and b≠1b \neq 1

When x=2x = 2 is substituted into the base x2−2x+1x^2 - 2x +1,

⇒ x2−2x+1=4−4+1=1x^2 - 2x +1 = 4 -4 +1 = 1

As the base is not allowed to be 1, the solution x=2x = 2 is rejected.

Substituting x=32=1.5x = \dfrac{3}{2} = 1.5 into the following expressions we get

x2−2x+1=2.25−3+1=0.25x^2 - 2x +1 = 2.25 - 3 +1 = 0.25

x2+x+1=2.25+1.5+1=4.75x^2 + x + 1 = 2.25 + 1.5 + 1 = 4.75

(The bases are positive and not equal to 1)

3x2−6x+7=6.75−9+7=4.753x^2 - 6x + 7 = 6.75 - 9 + 7 = 4.75

(The value is also positive)

Therefore, there is only 1 solution that exists, i.e., x=1.5x = 1.5

Answer: (2) 1


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