Question

how would I solve for x ?

( see message )

Answer 1

3^ sqrt x +4 = 27 ^sqrt x

Answer 2

the exponent of the LHS is

Answer 3

\[\sqrt{x} + 4\]

Answer 4

so what do you get when you change the base of the RHS

Answer 5

(3)^3(sqrtx)

Answer 6

yep, so now the bases on LHS and RHS are the same

\[\large3^{\sqrt x +4} = 3^{3\sqrt x} \]

taking the log to base three on both sides

\[\large\log_33^{\sqrt x +4} = \log_33^{3\sqrt x} \]

cancles the bases

\[\large{\sqrt x +4} = {3\sqrt x} \]

Answer 7

now just solve for \(x\)

Answer 8

thats the problem unkle, from that equation is it legal for this to happen

\[(\sqrt{x} + 4)^2 = (3\sqrt{x})^2\]

Answer 9

hmm, there is a much eaisier way

Answer 10

Please :)

Answer 11

**

√x+4 = 3√x

take away √x form both sides

Answer 12

\[\sqrt{x} + 4 = 3\sqrt{x} / \sqrt{x } ??\]

Answer 13

√x+4 -√x = 3√x -√x

4 = (3-1)√x

Answer 14

{the four wasent undrer the square root was it?}

Answer 15

yep

Answer 16

it was? or it wasn't?

Answer 17

it was not under the sqrt

Answer 18

ok good.

Answer 19

I'm a little confused about the RHS

Answer 20

oh you factored it ?

Answer 21

we had

3√x -√x right?

in other terms

3 B - B

to simplify

you have to realize that the terms are common

(3-1)B

we can combine the coefficients

and simplify

..B

Answer 22

\[(4 )^2= (2\sqrt{x})^2\]

Answer 23

16 = 4x

x = 4

is it correct ?

Answer 24

yeah x works out to equal 4

Answer 25

Thanks again unckle ! I shall do the checking :)

Answer 26

but you dont have to square the expression intill tha last step


4 = (3-1)√x
4 = 2√x
2 = √x
4 = x

Answer 27

Thanks for the tip :))