Logarithms

$9^{\log_{3}(4)} = 3^{2*\log_{3}(4)} = 3^{\log_{3}(4^{2})} = 3^{\log_{3}(16)} = 16$ .

Note that, in general, if $x = a^{\log_{a}(b)}$ then $\log_{a}(x) = \log_{a}(a^{\log_{a}(b)}) = \log_{a}(b)$ .

Now since the log function is one-to-one, we can conclude that $x = b$ .

$­{ 9 }^{ \log _{ 3 }{ 4 } }=x\\ \log _{ 3 }{ { 9 }^{ \log _{ 3 }{ 4 } } } =\log _{ 3 }{ x } \\ (\log _{ 3 }{ 4 } )log_{ 3 }{ 9 }=\log _{ 3 }{ x } \\ (\log _{ 3 }{ 4 } )(2)=\log _{ 3 }{ x } \\ { 3 }^{ \log _{ 3 }{ { 4 }^{ 2 } } }={ 3 }^{ \log _{ 3 }{ x } }\\ 16=x$

By rules of logarithms,

$x^{log_{a}b}$ can be written as $b^{log_{a}x}$ That makes things much more simpler.

Let $x = log_{3}(4)$ . Then we know that $3^{x} = 4$ . We want $9^{x}$ . So we can do this: $9^{log_{3}(4)} = 9^{x} = (3^{2})^{x}$ $(3^{2})^{x} = 3^{2x} = 3^{x} \times 3^{x}$
And since $3^{x} = 4$ , we're left with $4 \times 4 = 16$ .

Firstly, take a moment to understand this proof:

If $x = y^{\log_{b}a}$

Then, by taking log on both sides, we get $\log_{b}x = \log_{b}(y^{\log_{b}a}$ )

Using the identity $\log_{n}(m^x) = x\log_{n}m$

$\log_{b}x = ({\log_{b}a})\log_{b}y$

$\log_{b}x = (\log_{b}y){\log_{b}a}$

$\log_{b}x = \log_{b}(a^{\log_{b}y}$ )

Removing log, $x = a^{\log_{b}y}$

So therefore, combining the first and last statements, we get $y^{\log_{b}a} = a^{\log_{b}y}$

Moving on to the problem at hand:

Given $x = 9^{\log_{3}(4)}$

Using the above identity, we get

$x = 4^{\log_{3}(9)}$

$x = 4^2$

$x = 16$

${ 9 }^{ \log _{ 3 }{ 4 } }=({ 3 }^{ \log _{ 3 }{ 4 } })^{ 2 }=4^{ 2 }=16$

$Let\quad x=\log _{ 3 }{ 4 } \\ \\ So\quad 9^{ log_{ 3 }(4) }=9^{ x }\\ \\ But;\quad log_{ 3 }4\Leftrightarrow 3^{ x }=4\\ \\ \Rightarrow \quad 3^{ x }=4\quad (Square\quad both\quad sides)\\ \\ \Rightarrow \quad 3^{ 2x }=4^{ 2 }\\ \\ \Rightarrow \quad { \left( { 3 }^{ 2 } \right) }^{ x }=16\\ \\ \Rightarrow \quad 9^{ x }=16\\ \\ \therefore \quad 9^{ log_{ 3 }(4) }=16$

Let $x=9^{\log_{3}4}$ , we take logs on both sides to get log(x)=2log4, this implies that $x=16$ .

there is a rule...I don't have explanation now 9^log3(4) = 4^log3(9) you can actually rotate or swap the numbers, value remains the same.

Since a^logb c ( actually b is the base and a is input) = c^logb a( same as above) Therefore, the above statement can be rewritten as 4^log3 3^2=4^2=16

First, we can say 9 = 3^2.

Then, ${9}^{\log_{3}{(4)}} = {3}^{2\log_{3}{(4)}}$

Using the power property, ${2\log_{3}{(4)}}={\log_{3}{(16)}}$

Now, using the Logarithmic Rules of Exponents, ${a}^{\log_{a}{(b)}} = b$

Substituting all the values, we get ${3}^{\log_{3}{(16)}} = 16$

Finally, we can conclude the solution to be $\boxed{16}$ .

Does using Fermi's solution count? (i.e. I looked and guessed)

$9^{log_{3}(4)}$ = t,

taking log to the base e both sides,we have,

$[log_{3}(4)]*log 9$ = log (t).

$[log_{3}(4)]*2log(3)$ =log (t)

$\frac{log_{e}(4)}{log_{e}(3)}$ *2log(3)) = log (t)

$2*log(4)$ =log (t)

$log(4)^{2}$ =log (t)

log 16 =log (t)

hence.t=16,ans.

This is simpler imho:

$\begin{aligned} 9^{\log_3(4)} &=& (3\times 3)^{\log_3(4)}\\ &=&(3^{\log_3(4)})\times (3^{\log_3(4)})\\ &=& 4\times 4 \\ &=& 16 \end{aligned}$

Let be $log_{3} (4)=x$ , so $3^{x}=4$ and since $9=3^{2}$ it means that the result is the square of 4 so 16

P=9^log3 (4) take log for two sides for base 4 log4 (p)=log3 (4)×log4 (9)=log3 (9)=2 so p=4^2=16######

let A=9^log(3)4 let B=log(3)4

Then A=3^2B and 3^B=4 (by logarithm)

then A=16

Do not try to memorize formulas or theorems ; just do it yourself

$9^{log_3(4)} = (3^2)^{log_3(4)} = (3^{log_3(4)})^2 = 4^2 = 16$

you can rewrite that as log(base)9(of)x = log(base)3(of)4. Then, log(base)3^2(of)x can also be written as log(base)3(of)x^(1/2). Once the two expression on either side of the equal sine share the same base, you can set 4 equal to the square root of x, and therefore x is 16.

let>> x=9^log3(4)

log3(x)=log3(9^log3(4))= log3(4)* log3(9)=log3(4)*2= 2 log3(4) = log3(4^2)

so x=4^2= 16

9^{log3(4)}=4^log3(9)

= 4^ log3(3^2)
= 4^ 2log3(3)
= 4^ 2(1)
= 4^2 = 16

By luck babieesssssss ssssss

m^log b(n)=n^log b(m) hence question=4^log_3(9)=4^2

Lets suppose .... x= 9^log3 (4)....... x= 3^2*log3 (4).... x = 3^log3(4^2).....x=3^log3(16)......taking both sides log3.... log3(x)=log3[3^log3(16)]....... log3(x)=log3(16)log3(3).... log3(3)=1.....so log3(x)=log3(16).... by cancelling log3 on both sides we get.... x=16....hence.... 9^log3 (4)=16

Let 9^{log3 (4)} = x

Now, 3^2{log3 (4)} =x

Or,3{log3 (4^2)} =x

Now according to formula, we obtain, x= 4^2=16.

Log a(b) = log a^c(b^c) (proof at the bottom) Log 3(4) = log 9(16) 9^log_9(16) = 16

PROOF Log a^c(b^c) = log(b^c) / log(a^c) = c*log(b) /c*log(a) =log(b) / log(a) =log a(b)

X=9^(log₃4) log₃X=log₃4 log₃9=2 log₃4=log₃16 X=16

9^log3(4)

=> 3^2.log3(4)

m. log a= log a^m

therefore

3 ^log3(4^2)

if f(x)^log f(x) (X) = X ........... here f(x) is base in log

therefore

3^log3(4^2)= (4^2)=16 ....Ans

9^{log 3(4)}=3^2{log 3(4)}=3{log 3(4^2)}= 3{log 3(16)}= 16

9^log3(4)=3^log3(16) now go to basic knowledge, log3(16) means the exponent of 3 that will make it 16. suppose this exponent is x. so now we get 3^log3(16)=3^x=16. Bingo.

Say , y=9^[ log 4 to base 3 ].

Applying logarithms to base 3 on both sides ,

log y to base 3 = [ log 9 to base 3 ] * [ log 4 to base 3]. = 2 [ log 4 to base 3]. = [ log 4^2 to base 3 ]. = [ log 16 to base 3 ].

Comparing the arguments on both sides, y = 16.

Hence , 9^ [ log 4 to base 3 ] = 16.

9^log3(4) = 4^log3(9) = 4^2 = 16

you could also turn 3^(2log(3)(4)) into (3^log(3)(4)) (3^log(3)(4)) which intern equals 4 4=16

Let log_{3}(4)=y

4 = 3^{y}

squaring both sides

16 = 9^{y}

Hence 9^{log_{3}(4)} = 16

9^(log3(4)) = (3^2)^(log3(4)) = 3^(log3(4^2)) = 3^(log3(16)) = 16

9^(log3(4))=3^(2(log3(4)))=3^(log3(16))=16

9^(log3(4))=3^(2(log3(4)))=3^(log3(16))=16

Let X=given function.

than take log both side,

apply the basic principles of logarithms,

we get 16.