$$3^{4\log_{3} (x-1)+2}$$ can be simplified to

(a) 9(x-1)^4
(b) 6(x-1)^4
(c) 2+(x-1)^4
(d) 9+(x-1)^4
(e) 4x-2

Help please. I don't know how should i simplify this.

Thanks.

Question

$$3^{4\log_{3} (x-1)+2}$$ can be simplified to

(a) 9(x-1)^4
(b) 6(x-1)^4
(c) 2+(x-1)^4
(d) 9+(x-1)^4
(e) 4x-2

Help please. I don't know how should i simplify this.

Thanks.

satellite73 · Answer

it can't be "simplified" to anything
what does the actual question say?

rahulmr · Answer

That's all it says. I don't know what does it mean.

satellite73 · Answer

it doesn't mean anything
there is no such mathematical operation as "simplify" for one thing
for an other, all your possible answers seem to be exponential, not logs. 
perhaps it means "write in equivalent exponential form, but you would need an equation for that 
i

rahulmr · Answer

here, take a look at the image

satellite73 · Answer

oooh

satellite73 · Answer

you got a base of 3 there, didn't see it

rahulmr · Answer

yes

satellite73 · Answer

$$\huge 3^{4\log_3(x-1)+2}$$

satellite73 · Answer

now you can do something

satellite73 · Answer

first off, by one of the properties of the log, that 4 comes up as an exponent in $\log_3(x-1)$ 
so as a first step turn $$\huge 3^{4\log_3(x-1)+2}$$ in to $$\huge 3^{\log_3((x-1)^4)+2}$$

rahulmr · Answer

ok

satellite73 · Answer

then using another law of exponents, that $$b^xb^y=b^{x+y}$$ you can make this $$\huge3^3\times 3^{\log_3((x-1)^4)}$$

satellite73 · Answer

oops should be $3^2$ there, not $3^3$

rahulmr · Answer

i was going to say this

satellite73 · Answer

$$\huge9\times 3^{\log_3((x-1)^4)}$$

satellite73 · Answer

and since logs are the inverse of the exponential, you have $$3^{\log_3((x-1)^4)}=(x-1)^4$$

satellite73 · Answer

so final answer would be $$\huge 9(x-1)^4$$

rahulmr · Answer

ohh. now i get it. Thanks for the help. I was confused because of the 3.

satellite73 · Answer

yw