Question

If log3x+ log3 4=2 then x=
1) 9/4
2) 4/9
3) 9
4) 0
Please solve this MCQ​

Answer 1

Given Question :-

$\rm \: If \: log_{3}(x) + log_{3}(4) = 2, \: then \: x =$

$\rm \: \: \: \: \: \: \: 1) \: \dfrac{9}{4}$

$\rm \: \: \: \: \: \: \: 2) \: \dfrac{4}{9}$

$\rm \: \: \: \: \: \: \: 3) \: 9$

$\rm \: \: \: \: \: \: \: 4) \: 0$

$\large\underline{\sf{Solution-}}$

Given Logarithmic equation is

$\rm \: log_{3}(x) + log_{3}(4) = 2$

We know,

$\boxed{\sf{ \: log_{x}(a) + log_{x}(b) \: = \: log_{x}(ab) \: }}$

So, using this identity, we get

$\rm \: log_{3}(4x) = 2$

We know, that

$\boxed{\sf{ \: log_{x}(y) \: = \: z \: \: \rm\implies \:x = {y}^{z} \: \: }}$

So, using this identity, we get

$\rm \: 4x = {3}^{2}$

$\rm \: 4x = 9$

$\rm\implies \:x = \dfrac{9}{4}$

$\rm\implies \:Option \: 1) \: is \: correct.$

Verification :-

Consider, LHS

$\rm \: log_{3}(x) + log_{3}(4)$

On substituting the value of x, we get

$\rm \: = \: log_{3}\bigg( \dfrac{9}{4} \bigg) + log_{3}(4)$

$\rm \: = \: log_{3}\bigg( \dfrac{9}{4} \times 4\bigg)$

$\rm \: = \: log_{3}(9)$

$\rm \: = \: log_{3}( {3}^{2} )$

$\rm \: = \: 2$

Hence, Verified

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Additional Information

$\boxed{\sf{ \: log_{x}(a) + log_{x}(b) \: = \: log_{x}(ab) \: }}$

$\boxed{\sf{ \: log_{x}(a) - log_{x}(b) \: = \: log_{x}\bigg( \frac{a}{b} \bigg) \: }}$

$\boxed{\sf{ \: log_{x}( {a}^{b} ) \: = \: b \: log_{x}(a) \: \: }}$

$\boxed{\sf{ \: \: log_{x}(x) \: = \: 1 \: }}$

$\boxed{\sf{ \: \: log_{x}( {x}^{y} ) \: = \: y \: \: }}$

$\boxed{\sf{ \: \: log_{ {x}^{z} }( {x}^{y} ) \: = \: \frac{y}{z} \: \: }}$

$\boxed{\sf{ \: \: {a}^{ log_{a}(x) } \: = \: x \: }}$

$\boxed{\sf{ \: \: {a}^{y \: log_{a}(x ) } \: = \: {x}^{y} \: }}$

Answer 2

Answer:

We know that,

logₙ (y) = z => x = y²

So,

4x =3²

4x = 9

=> x = 9/4

So, option 1.) is correct

Now, Verification

log₃ (9/4) + log₃ (4)

log₃ (9/4×4)

log₃ (9)

log₃ (3²)

= 2

Hence, Verified