Question

log (base 6) (x+3) + log (base 6) (x+4)=1
**open question to see the equation properly.

Answer 1

\[\log_{6}(x+3)+\log_{6}(x+4)=1 \]

Answer 2

Basically, let both sides be exponents of \(6\).

Answer 3

\[6^{\log_{6}(x+3)+\log_{6}(x+4)}=6^1 \]

Answer 4

what does that do though?

Answer 5

It will get rid of the \(\log_6\) if you know your properties of exponents.

Answer 6

Do you know\[
a^{b+c} =\ldots
\]

Answer 7

would that make it (x+3)(x+4)=6?

Answer 8

...




Yes

Answer 9

alright one sec let me finish working i tout and see if it works out

Answer 10

it worked thanks!

Answer 11

\[9^x+4 \times 3^x-3=0\] @wio could you help me with this problem too please?

Answer 12

\[9^x+4 \times 3^x-3=0\] Well, let \(u=3^x\).

Answer 13

\[
u^2+4u-3=0
\]

Answer 14

Our \(x\) solutions correspond to our \(u\) solutions.

Answer 15

I mean, we could have alternatively substituted \(x =\log_3(u)\).

Answer 16

the answer given is in log form so does that mean i use the alternative?

Answer 17

\[\log(a+b)=\log(ab)\]\[\log_6(x+3) +\log_6(x+4) =1\]\[\log_6[(x+3)(x+4)] =1\]\[(x+3)(x+4)=6\]

Answer 18

That's how I tried solving it.

Answer 19

that's kind of what i got

Answer 20

\[\log(a+b)=\log(ab)\]???

Answer 21

No, what it means is you solve for \(u\). You use quadratic formula.

Answer 22

Then, ones you have your two values for \(u\)... suppose they are \(u_1,u_2\), then you solve for \(x\)\[
3^{x_1}=u_1\\
3^{x_2}=u_2\\
\]

Answer 23

In short, find your \(u\) values, and find your \(x\) values using: \(x = \log_3(u)\).

Answer 24

\[
u^2+4u-3=0
\]So :\[
u = \frac{-4\pm\sqrt{4^2-4(1)(-3)}}{2(1)}
\]And \[
x = \log_3\left(\frac{-4\pm\sqrt{4^2-4(1)(-3)}}{2(1)}\right)
\]

Answer 25

\wait when you solve that what ar eyou supposed to get at the end?

Answer 26

@wio