Question

Logarithm question

Answer 1

\(\large \color{black}{\begin{align}& \text{solve}\hspace{.33em}\\~\\& \log_{2}(9-2^{x})=10^{\log_{10}(3-x)}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

@freegirl11112

Answer 3

@freckles

Answer 4

hi I dot know this yet sorry

Answer 5

mistyped ur name was calling frekels

Answer 6

\[\log_2(9-2^x)=10^{\log(3-x)} \\ \log_2(9-2^x)=3-x \\ 2^{3-x}=9-2^x \\ 2^{3-x}+2^x=9 \\ 2^3 2^{-x}+2^x=9 \\ 8 \cdot 2^{-x}+2^x=9 \\ \text{ multiply both sides by } 2^x \\ 8 +2^{2x}=9 \cdot 2^{x}\]

Answer 7

i didn't understand the second step \(\log_2(9-2^x)=3-x\)

Answer 8

what happened to 10^log ?? :o

Answer 9

sorry log_10(x) and 10^x are inverses

Answer 10

\[10^{\log_{10}(3-x)}=3-x\]
when 3-x>0

Answer 11

mhm oh okay thanks:=)

Answer 12

and I'm so sorry but I have to go
someone is waiting on me

Answer 13

hey @mathmath333 were you able to finish

Answer 14

you have a quadratic in terms of 2^x

Answer 15

yep o got it.