Are you sure this can be done

$4^m = 8$

$2^{2m} = 2^3$

$2m = 3$

$9^m = 3 ^{2m} = 3^3 = 27$

$3^n = 5$

$9^n = 3^{2n} = (3^n)*(3^n) = 5*5 = 25$

$So,$

$9^{n+m} = (9^n)*(9^m) = 25*27 = 675$

Finding $9^{n+m}$ is the same as finding the product of $9^n$ and $9^m$ .

If we consider the expression $3^n=5$ , we see that raising each side to the second power gives us:

$3^{2\cdot n}=5^2$

$9^n=25$ .

Now that we know the value of $9^n$ , we focus on finding the value of $9^m$ . To do so, we must solve for $m$ , which can be done as follows:

$4^m=8$

$m\log_24=\log_28$

$m=\frac{\log_28}{\log_24}$

$m=\frac{3}{2}$

Now that we know $m$ we can compute $9^m$ .

$9^m=\sqrt{9^3}$

$9^m=\sqrt{729}$

$9^m=27$ .

Thus, $9^{n+m}=9^n9^m=25\cdot27=\boxed{675}$ .

$8=2^3$

$4^m=(2^2)^m=2^{2m}$

$4^m=8 \Rightarrow 2^{2m}=2^3 \Rightarrow 2m=3$

$3^n=5 \Rightarrow (3^n)^2=5^2 \Rightarrow 3^{2n}=25$

$9^{n+m} = (3^2)^{n+m} = 3^{2n+2m} = 3^{2n} \times 3^{2m} = 25 \times 3^3 = \fbox{675}$

$3^{n} = 5$ $4^{m} = 2^{2*m} = 8$ so that $m = 3/2$

$9^{(n+3/2)} = 3^{2*(3/2+n)} = 3^{3}+(3^{n})^{2} = 3^3+5^2 = 675$

$3^{n}$ = $5$ ,

$9^{n}$ = $(3^{n})^{2}$ = $25$ .

$(2^{2})^{m}$ = $2^{3}$ ,

m= $\frac{3}{2}$ , so ( $9^{m}$ )= $27$

Hence $9^{n+m}$ = $25 \times 27$ = $675$

${3^n}={5}$

Taking *ln on both sides* ${ln3^n}={ln5}$

we get ${nln3}={ln5}$

and ${n}=\frac{ln5}{ln3}$

${4^m}={2}^{2m}={2^3}$

${2^2m}={2^3}$

applying the laws of indices, ${2m}={3}$

and ${m}=\frac{3}{2}$

i.e. $9^{\frac{ln5}{ln3}+\frac{3}{2}}$

which gives $\boxed{675}$

$3^{n}=5 \implies n=\log_{3} 5$ and $4^{m}=8 \implies n=\log_{4} 8$ .

$\log_{4} 8 = \log_{4} 4 \times 2 = \log_{4} 4 + \log_{4} 2 = 1 + \frac{1}{2} = \frac{3}{2}$

$\implies 9^{n+m} = 9^{\log_{3} 5 + \frac{3}{2}} = 9^{\frac{3}{2}} \times 9^{\log_{3} 5}$

$9^{\log_{3} 5} = \left(3^{2}\right)^{\log_{3} 5} = 3^{2 \times \log_{3} 5} = 3^{\log_{3} 25} = 25$

$\implies 9^{n+m} = 9^{\frac{3}{2}} \times 25 = 27 \times 25 = 675$

$1.$ $4^m=8$

$\implies2^{2m}=2^3$

$\implies 2m=3$

$\implies m=\frac{3}{2}$

$2.$ $3^n=5$

$\implies n=log_35$

$Now$ , $9^{n+m}$

$\implies 3^{2{n+m}}$

$\implies 3^{2{log_35+log_33^{3/2}}}$

$\implies 3^{log_3{5.3^{3/2}}^{2}}$

$\implies ({5.3^{3/2}})^2$

$\implies 25.3^3$

$\implies 25.27$

$\implies 675$

First of all, let us modify the original problem a bit:

$9^{n+m}=3^{2n+2m}=3^{2n}3^{2m}$

Noting that $3^{2n}=(3^{n})^{2}=25$ , we can see that the original expression becomes $25\cdot3^{2m}$ . This means that all we need to do is to find $m$ .

Let us now take a look at $4^{m}=8$ . Note that this is equivalent to $2^{2m}=2^{3}$ . Furthermore, by using the fact that $log_{x}x=1$ , we get

$2^{2m}=2^{3} \Rightarrow \log_{2}2^{2m}=\log_{2}2^{3} \Rightarrow 2m=3 \Rightarrow m=\frac{3}{2}$

Thus, if we go back to the original expression we will finally get

$25\cdot 3^{2m}=25\cdot 3^{3}=25\cdot 27=675$ .

$3^{n}$ =5,means n= $\log _{ 3 }{ 5 }$

$4^{m}$ =8,m=4 ${\frac { 3 }{ 2 } }$

$9^{n+m}=9^{ \log _{ 3 }{ 5 } + \frac { 3 }{ 2 }}$

=675(rounded)

So here it goes. Let us break 9 as the square of 3. The statement 9 to the power (n+m) then becomes 3 to the power (2n+2m). This implies 9 to the power 2n is multiplied by 9 to the power 2m. What we need now are the values of m and n. Look carefully that 4 to the power m is actually 2 to the power 2m and 8 is 2 cubed The bases being equal, we have 2m equal to 3. This gives m =3/2 and 2m = 3. Moreover 3 to the power 2n is actually 3 to the power n multiplied by itself. This implies 5 multiplied by itself(for obvious reasons!). Thus, we are done with finding the required values if not the individual values of m and n respectively! Now 3 to the power (2n+2m) can be written as 3 to the power 2n multiplied by 3 to the power 2m. We have got the value 3 to the pwer 2n which is fortunately the 5 squared and 3 to the power 2m is actually 3 cubed. Thus, here it is. 5 squared multiplied by 3 cubed gives you 675. Solved!

• 3^{n}=5; 4^{m}=8 => 2^{2m}=8 => 2m=3

• 9^{n+m}=9^{n} * 9^{m} = 3^{2n} * 3^{2m} = (3^{n})^{2} * 3^{2m} = 5^{2} * 3^{3} = 25*27= 675

The question is $9^{n+m}$ , so we can change it as $9^{n} \times 9^{m}$ . First, let's find $9^{n}$ . $9^{n} = 9^{^{3}log5} = 9^{2 \times ^{9}log5} = (9^{^{9}log5})^{2} = 5^{2} = 25$ Now, for the $9^{m}$ $9^{m} = 9^{^{4}log8} = 9^{\frac{3}{2} \times ^{8}log8} = 9^{\frac{3}{2}} = 27$ So, $9^{n} \times 9^{m} = 25 \times 27 = 675$

Well there are many ways of approaching this problem. Im my way, there are some basic principles that one should know in order to solve this question.

a^{b+c) = a^b + a^c

Also, you should know that you can substitute $9$ for $3^2$

So the question asks what is the value of $9^{m+n} \implies 3^{2(n+m)} \implies 3^{2n} \times 3^{2m}$

Now its rather easy to solve for 'm' (from the equation $4^m = 8$ ) 'm' is equal to 3/2

And we don't actually have solve for 'm', rather because we know that $3^n = 5$ , we know that $3^{2n} ={25}$ because we square both sides of the equation.

$3^{2m} \implies 3^{2 (3/2)} \implies 27$

So we want to find $9^{m+n} \implies 3^{2n} \times 3^{2m} \implies 25 \times 27 \implies 675$

1. 4^m = 8 2^2m = 2^3 2m=3 m=3/2

so, 9^n x 9^m = 3^n x 3^n x 3^2m = 5 x 5 x 27 = 675

4^m=2^2m=8=2^3 => m=3/2.

9^(n+m)=9^n.9^m

9^n=3^2n=5^2=25.

9^m=9^(3/2)=27.

25.27=675

We have $3^n = 5$ , so $9^n = = \left( 3^2 \right)^n = \left( 3^n \right)^2 = 25$ . Also, $4^m = 8$ , so $m = \dfrac {3}{2}$ . But note that $9^{n+m} = 9^n \cdot 9^m = 25 \cdot 9^{\frac{3}{2}} = 25 \cdot 27 = \boxed {675}.$

9^(n+m) can be written as (3^n)^2 X 9^m which is equal to 5^2 x 9^m since 3^n=5. Hence, result becomes 25 x 9^m. Now since 4^m=8; m=2x1/2 (sq.root2); hence 9^m = 81. Thus result is 25 x 27 i.e. 675.

If 3^n=5, 9^n=25 (squaring on both sides) - (1). And, since 4^m=8, 2^(2m)=2^(3) (since 4=2^2 & 8=2^3). Therefore, 2m=3 (f bases are same, exponents can be equated) =>m=3/2. We want 9^(n+m)= (9^n) x (9^m) = (25) x (9^(3/2)) = (25) x {[9^(1/2)]^3}
= (25) x {[3]^3} = (25) x {27} = (25) x {25+2} = 25x25 + 25x2 = 625 + 50
= 675.

9^n+m= 9^n * 9^m=(3^n)^2 * (3^2)^m since 4^m=8 therefore 2^2m=2^3 hence 2m=3 m=1.5 5^2 * 3^(2*1.5)=25 * 27=675

3^n = 5 n(log 3) = (log 5) n= 1.46

4^m=8 2^2m=2^3 2m=3 m=1.5

n+m= 1.46+1.5 = 2.96

9^2.96 =675

Although we cannot determine n, we can determine m. Note that:

4 = $4^{1}$ < 4 + 4 = 8 < 8 + 8 = $4^{2 }$

So a reasonable guess for m is 1.5 knowing how a quadratic graph behaves. In fact $4 ^ {1.5}$ = 8.

Now note that:

$9^{m+n}$ = $9^{m} \times 9^{n}$ = $9^{1.5} \times 9^{n}$ = $27 \times 9^{n}$ (*)

Remark that:

$9^{n} = (3^{2})^{n} = (3 ^{n})^{2} = 5^{2} = 25$

Subbing this into (*) gives us that the answer to this question is $25 \times 27 = 675$

De $4^m = 8$ temos que:

$(2^2)^m = 8 \Rightarrow 2^{2m} = 2^3 \Rightarrow 2m = 3 \Rightarrow \boxed{m = \frac{3}{2}}$

Segue que,

$9^{n + m} \Rightarrow 3^{2n + 2m} \Rightarrow 3^{2n} \cdot 3^{2m} \Rightarrow 3^n \cdot 3^n \cdot 3^{3} \Rightarrow 5 \cdot 5 \cdot 27 \Rightarrow \boxed{\boxed{675}}$

$3^n = 5 ~~~ \Longrightarrow ~~~n= \log_3 5$

$~$

$4^m=8 ~~~ \Longrightarrow ~~~m = \log_4 8$

$~$

$\therefore ~~~9^{m+n} ~~~=~~~ 9^m \cdot 9^n ~~~=~~~ 9^{\log_4 8} \cdot 9^{\log_3 5}$

$~~~~~=(3^2)^{\log_4 8} \cdot (3^2)^{\log_3 5}~=3^{2\log_4 8} \cdot 3^{2\log_3 5}$

$~~~~~=3^{\log_4 8^2} \cdot 3^{\log_3 5^2}~=3^{2\log_4 64} \cdot 3^{\log_3 25}$

$~~~~~=3^{\log_4 4^3} \cdot 3^{\log_3 25} = 3^3 \times 25 = 27 \times 25 = \fbox{675}$

we've $3^n = 5$ , and $4^m = 8$ , so:

$n = log_3(5)$ and $m = log_4(8)$ ,

than :

$n + m = log_3(5) + log_4(8) = log_9(675) = \frac{log(675)}{log(9)}$

also we've:

$x^{(\frac{1}{log(x)})} = e$ because by add log to both sides we've:

$log(x^{(\frac{1}{log(x)})}) = log(e)$

$(\frac{1}{log(x)}) \times log(x) = 1$

$1 = 1$

so:

$9^{(log_9(675))} = 9^{( \frac{log(675)}{log(9)})} = (9^{(\frac{1}{log(9)})})^{log(675)} = e^{log(675)} = 675$