Question

Solve for x:

5^x * 2^(x-1) = 100

Answer 1

@Mertsj

Answer 2

guess last answer didn't help
want to try again?

Answer 3

I tried but I wasn't able to cancel out the logs to get the x value

Answer 4

there is no cancelling
it is a raft of algebra

Answer 5

Could you show me how it's done. The first part I understand

Answer 6

did you understand how to get
\[x\ln(5)+(x-1)\ln(2)=\ln(100)\]?

Answer 7

Yes

Answer 8

k then remember that although \(\ln(x)\) is a function, \(\ln(2), \ln(5),\ln(100)\) are numbers
i.e. constants

Answer 9

so it will be like solving
\[2a+b(x-1)=c\] for \(x\)
i can walk you through it if you like, but it is algebra from here on in

Answer 10

oops i meant
it will be like solving
\[ax+b(x-1)=c\] for \(x\)

Answer 11

Could u please go through it

Answer 12

\[ax+b(x-1)=c\\
ax+bx-b=c\\
ax+bx=c+b\\
(a+b)x=c+b\\
x=\frac{c+b}{a+b}\]

Answer 13

So we plug in the values to get x?

Answer 14

repeat with
\[\ln(2)x+\ln(5)(x-1)=\ln(100)\] it is identical

Answer 15

as a matter of fact, now that we have it with \(a,b,c\) we can go right to
\[x=\frac{\ln(100)+\ln(5)}{\ln(2)+\ln(5)}\]

Answer 16

if you want a decimal, use a calculator

Answer 17

Thank you so much :)

Answer 18

also if you like you could use properties of the log to rewrite as
\[x=\frac{\ln(500)}{\ln(10)}\]

Answer 19

yw

Answer 20

Sorry to bother you but I've you plug in the x value (2.6989) in the original equation it doesn't equal 100

Answer 21

@satellite73

Answer 22

I found the error it should be In 200 instead of In 500

Answer 23

Anyways thanks for your help. Now I understand how its done