Question

Solve 9*3^x=5

Answer 1

try dividing 9 on both side to isolate the 3^x part

Answer 2

then take ln( ) of both sides

Answer 3

you mean this right? \[9*3^{x} = 5\]

Answer 4

Just treat 3^x as a variable and isolate that. So you have nine times 3. you need to do the opposite which is division.

\[3^{x} = \frac{ 9 }{ 5 }\]

then you have 3^x by itself. you need to apply logs to solve this. what you would do is take the log of both sides.

I'll stop there.

Answer 5

that should be 5/9

Answer 6

Or you can rewrite 3 as 3^2 first if you don't like fractions.

Answer 7

i mean 9 as 3^2

Answer 8

\[\rm 9= 3 \cdot 3 =3^2\]

Answer 9

So I have \[3^2*3^x=5\]

Answer 10

you could use law of exponents if you go @Nnesha 's route which says x^a*x^b=x^(a+b)

Answer 11

so 3^2 * 3^x =?

Answer 12

Wouldn't that be 3^(2+x)=5?

Answer 13

that's cool
you still can't get past applying a log to both sides
I like to use ln( )

Answer 14

Alright, how would you do with the natural log?

Answer 15

\[9\cdot \:3^x=5\quad :\quad x=-\frac{\ln \left(\frac{9}{5}\right)}{\ln \left(3\right)}\quad \left(\mathrm{Decimal:\quad }x=-0.53503\right)\]

Answer 16

you do ln( ) of both sides
this allows you to bring the power down

@DecentNabeel It would be cool if we can lead him to the answer instead of just giving him the answer

Answer 17

\[3^{2+x}=5 \\ \ln(3^{2+x})=\ln(5) \\ \text{ use power rule: } \ln(a^r)=r \ln(a)\]

Answer 18

I guess what I'm confused is the answer that the teacher gave of x = log3
(5) − 2

Answer 19

is that log base 3?

Answer 20

Yes

Answer 21

if so then take log base 3 of both sides instead of ln( )

Answer 22

you will still use power rule though

Answer 23

@myininaya happy

Answer 24

@myininaya could you show what that looks like

Answer 25

instead of seeing the ln( ) on both sides
write \[\log_3( )\] on both sides

Answer 26

you know where I had written ln( )?

Answer 27

Oh never mind I got it

Answer 28

except for the -2 part

Answer 29

\[9\cdot \:3^x=5\]
\[\mathrm{Divide\:both\:sides\:by\:}9\]
\[3^x=\frac{5}{9}\]
\[\mathrm{Use\:the\:following\:exponent\:property}:\quad \frac{1}{a^n}=a^{-n}\]

\[\frac{5}{9}=\left(5\right)9^{-1}\]
\[3^x=5\cdot \:9^{-1}\]

\[\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)\]

Answer 30

@DecentNabeel thanks but that doesn't solve for x

Answer 31

\[\ln \left(3^x\right)=\ln \left(5\cdot \:9^{-1}\right)\]

\[\log _a\left(x^b\right)=b\cdot \log _a\left(x\right):\quad \ln \left(3^x\right)=x\ln \left(3\right)\]
\[x\ln \left(3\right)=\ln \left(5\cdot \:9^{-1}\right)\]

Answer 32

\[\mathrm{Solve\:}\:x\ln \left(3\right)=\ln \left(5\cdot \:9^{-1}\right):\quad x=-\frac{\ln \left(\frac{9}{5}\right)}{\ln \left(3\right)}\]

Answer 33

\[3^{x+2}=5 \\ \text{ take } \log_3( ) \text{ of both sides } \\ \log_3(3^{x+2})=\log(5) \\\]

You can use power rule here... the one I mentioned above where you can bring down the exponent on the thing inside the log

Answer 34

\[x\ln \left(3\right)=\ln \left(5\cdot \:9^{-1}\right)\]
\[\mathrm{Divide\:both\:sides\:by\:}\ln \left(3\right)\]

\[\frac{x\ln \left(3\right)}{\ln \left(3\right)}=\frac{\ln \left(5\cdot \:9^{-1}\right)}{\ln \left(3\right)}\]
\[x=-\frac{\ln \left(\frac{9}{5}\right)}{\ln \left(3\right)}\]

Answer 35

@smileyxl3 gotit

Answer 36

that one side forgot to write the subscript 3 (the base I mean)

\[3^{x+2}=5 \\ \text{ take } \log_3( ) \text{ of both sides } \\ \log_3(3^{x+2})=\log_3(5) \\\]
\[log_a(a)=1\]
\[\text{ for any } a \text{ that is in } (0,1) \text{ union } (1,\infty)
\]
can you try to use power rule on the left hand side and then use that last thing I wrote as well