Question

If i have a question like 3(5^x+1) = 15
I ended up with 15^x-1 = 15
If I then drop the bases since they are equal does the 15 on the rights exponent end up as 1?

Answer 1

is your question is\[3(5^x+1)=15\] ?

Answer 2

yes

Answer 3

then you should proceed like 3(5^x+1)=15
5^x+1=15/3;
5^x+1=5
5^x=4
\[x=\log_{5}{4}\]

Answer 4

i don't know how did u get 15^x-1=15???

Answer 5

the answer in the book is 0. I wrote that wrong it was still a +.

Answer 6

if u need zero then the question should look like:\[3*5^{(x+1)}=15\]

Answer 7

check it...otherwise you can't get 0

Answer 8

Yes, im sorry, how did you write that?

Answer 9

\[3*5^{(x+1)}=15:\]\[5^{x+1}=\frac{15}{3}\]\[5^{x+1}=5\]\[x+1=1;\]x=0

Answer 10

got it?

Answer 11

I couldn't just multiply the 3 with the 5 to get 15?

Answer 12

no..

Answer 13

its not 5 its 5^(x+1)

Answer 14

for example \[3*5^2\neq (3*5)^2\]

Answer 15

Order of operations, you must deal with the exponent before you can start multiplying.

Answer 16

i think you need a little study on the properties of indices...

Answer 17

I do. 10 years from math and i've lost most of these things. im trying to get th em back through trial and error, since I dont have time to redo earlier classes.

Answer 18

okay..carry on...don't give up...someone is always here to help you.

Answer 19

I appreciate it greatly!

Answer 20

If i may, dipankarstudy just made x a value so that the 5^(x+1) would equal 5, and thus it would be easy to manipulate the 3*5. If you can see he was able to quickly solve for x because of that, it makes the problem much easier.