Question

solve the following exponential equation
9^x-14*3^x=-49
(a)What is the exact solution.

(b)what is the decimal approximation for the solution.

Answer 1

\[9^x-14(3^x)+49=0\\
3^{2x}-14(3^x)+49=0\\
{\rm let}\quad u=3^x
\]

Answer 2

The equation is

\[\Large 9^x - 14*3^x = -49\]

right?

Answer 3

becomes a quadratic

Answer 4

if so, then use electrokid's method to get

u^2 - 14u + 49 = 0

(u - 7)^2 = 0

u - 7 = 0

u = 7

3^x = 7

Answer 5

now rewrite the equation in-terms of "u" variable

Answer 6

keep going to solve for x

Answer 7

right,
so what you end up with is:
\[3^x=7\]
how do we bring the "x" down from the exponent?

Answer 8

use the log and make it xlog3=7?

Answer 9

you mean,
\[x\log(3)=\log(7)\]

Answer 10

yes!

Answer 11

now divide both sides by log(3) to isolate x

Answer 12

and voila!

Answer 13

so the exact solution would be x= log of 7/ log of 3

Answer 14

yep.
you might also have ..
\[x=\log_3(7)\]

Answer 15

okay thanks!

Answer 16

**using change of base rule**

Answer 17

I have a quick question would 4log(base5)(sqareroot of (9x-4)-log (base5) (2/x)+log(base 5) (2) would have the simplified form of log (base5)((9x-4)^2)/(2x-3))(2)

Answer 18

where are you getting 2x-3 ?

Answer 19

I really dont remember. but it feel like it was wrong. can you help me with it?

Answer 20

ok each log is assumed to be base 5

Answer 21

okay

Answer 22

4*log(sqrt(9x-4)) - log(2/x) + log(2)

4*log((9x-4)^(1/2)) - log(2/x) + log(2)

log((9x-4)^((1/2)*4)) - log(2/x) + log(2)

log((9x-4)^2) - log(2/x) + log(2)

log((9x-4)^2) + log(2) - log(2/x)

log(2(9x-4)^2) - log(2/x)

log( [2(9x-4)^2] / (2/x) )

log( [2x(9x-4)^2] / 2 )

log( x(9x-4)^2 )

Answer 23

thank super much! i can see the steps and understand it!

Answer 24

ok glad it's making sense now