Question

solve the equation by using logarithm
7^x-1= 5^2x+1 leave answer with log
please show all work

Answer 1

(x-1)log7=(2x+1)log5

Answer 2

divid a log from either side and solve for x

Answer 3

apply log on both sides
\[\log7^{x-1}=\log(5^{2x}+1)\]
\[\log7^{x-1}=\log5^{2x} * log1\]

\[\ (x-1)log7=0\]
x =1

Answer 4

wait, is the 2x+1 in the exponent of 5 or is it just 2x?

Answer 5

2x+1 is the exponents of 5

Answer 6

oh, then do what annon said.

Answer 7

can u do it?

Answer 8

yes, but I prefer you do it.

Answer 9

hold the phone

Answer 10

it is not the case that
\[log(5^{2x}+1)=log(5^{2x})\times log(1)\]

Answer 11

is u studying? dhatraditya

Answer 12

that would make everything 0!

Answer 13

satellite, I had the same thought, but apparently it is \[5^{2x+1}\]

Answer 14

doesn't matter it is never the case

Answer 15

no i'm not studying. Im here to help

Answer 16

\[7^{x-1}=5^{2x+1}\] yes?

Answer 17

yup

Answer 18

or (x-1) log 7 = (2x+1) log 5

Answer 19

or 1.2(x-1) = 2x+1
solve for x

Answer 20

\[ln(7^{x-1})=ln(5^{2x+1})\]
\[(x-1)ln(7)=(2x+1)ln(5)\]

Answer 21

\[ln(7)x-ln(7)=2ln(5)x+ln(5)\]

Answer 22

I don't know anything about log. I read in 9th grade. and these problems are from my college now class

Answer 23

\[ln(7)x-2ln(5)x=ln(5)+ln(7)\]

Answer 24

\[x(ln(7)-2ln(5))=ln(5)+ln(7)\]

Answer 25

and finally
\[x=\frac{ln(5)+ln(7)}{ln(7)-2ln(5)}\]

Answer 26

only log step was to get the variable out of the exponent. all else was algebra

Answer 27

you can write the answer in different ways using propertties of the logs

Answer 28

for example
\[ln(5)+ln(7)=ln(35)\] but i would just leave it

Answer 29

halimabegum wrote "I don't know anything about log. I read in 9th grade. and these problems are from my college now class"

Thats hardly an excuse.

Answer 30

okay!

Answer 31

problems like these are a pain where you have different bases. you can see all the algebra involved