+0

# If $\left(\sqrt[4]{11}\right)^{3x-3}=\frac{1}{5}$, what is the value of $\left(\sqrt[4]{11}\right)^{6x+2}$? Express your answer as a fractio

+1
536
2
+598

If $$\left(\sqrt[4]{11}\right)^{3x-3}=\frac{1}{5}$$, what is the value of $$\left(\sqrt[4]{11}\right)^{6x+2}$$? Express your answer as a fraction.

off-topic
Sep 16, 2017
edited by michaelcai  Sep 16, 2017

#1
+3
+2

I hope this helps ( i took a snapshot from word, cause LaTeX gave me cancer ).

Sep 16, 2017
#2
+98173
+1

Thanks, Dr Dros!!!

Here's another method without resorting to logs

[(11)^ (1/4)] ^(3x -3)  = 1/5     implies that

[  11 ^(x - 1) ] ^ (3/4)   = 1/5       take both sides to the 4/3 power

[11 ^(x - 1) ]   =  (1/5)^(4/3)

[11^x ] / 11  =   (1/5)^(4/3)

11^x  =  11* (1/5)^(4/3)

So.....  11^(6x)  =  (11^x)^6  =  [ 11 * (1/5)^(4/3) ]^6  =  [ 11^6] * [ 1/5]^8

So..... [11^(1/4)]^(6x + 2)  =   [ 11^(6x + 2) ] ^(1/4)  =  [ 11 ^(6x) * 11^2] ^ (1/4)  =

[ 11^6 * (1/5)^8 ] ^(1/4) *  11^(1/2)   =

[ 11^(6/4)] * [(1/5)^8]^(1/4) * 11^(1/2)  =

[11 ^ (3/2) * [ (1/5) ^2] * 11^(1/2) =

[ 11 ^2]  *  (1 / 25)  =

121  / 25

Sep 16, 2017