Question

logarithmic expressions: solve to the nearest thousandth..
\[5^{1 + x} = 2^{1-x}\]

Answer 1

start with
\[(x+1)\ln(5)=(1-x)\ln(2)\] and then do a bunch of algebra recalling that \(\ln(5)\) and \(\ln(2)\) are just some constants

Answer 2

i can walk you through it if you like, but it is really algebra to get \(x\) by itself on one side of the equal sign

Answer 3

yes please, cause i got through first two equations of the question its just this one i have left.

Answer 4

\[(x+1)\ln(5)=(1-x)\ln(2)\]
\[x\ln(5)+\ln(5)=\ln(2)-x\ln(2)\]
\[x\ln(5)+x\ln(2)=\ln(2)-\ln(5)\]
\[x(\ln(5)+\ln(2))=\ln(2)-\ln(5)\]
\[x=\frac{\ln(2)-\ln(5)}{\ln(5)+\ln(2)}\]

Answer 5

then a calculator, because i have no idea what this is

Answer 6

LOL okay. wait but if you bring (ln(5)+ln(2)) to the right side doesn't it become (ln(5)-ln(2)) ??

Answer 7

oh... wait... nvm!

Answer 8

not sure what you are asking
steps are
1) distributive property
2) add \(x\ln(2)\) to both sides
3) subtract \(\ln(5)\) from both sides
4) factor the \(x\)
5) divide

Answer 9

lol sorry i confused myself for a second there.

Answer 10

easy to do but thing of \(\ln(5)=a\) and \(\ln(2)=b\) they are just constants

Answer 11

*think