Question

log8(x+1)=1−log8(x+8)

Answer 1

\[\log_{8} (x+1) = 1-\log_{8} (x+8)\]

Answer 2

Exponentiate both sides with a base 8, and see what you can do with that.\[\huge 8^{\log_8(x+1)}=8^{1-\log_8(x+8)}\]

Answer 3

can u show me the steps

Answer 4

No, I want you to actually put effort in. Remember that \[\huge a^{\log_a b}=b\] and \[\huge a^{b+c}=a^ba^c\]

Answer 5

i dont get it

Answer 6

Use those rules I gave to simplify the expression I wrote above.

Answer 7

so its eqial to x plus 1?

Answer 8

That's the left hand side, yes. Can you figure out the right hand side?

Answer 9

8 to the power of 1?

Answer 10

Not quite. The exponent has more than just a 1 in it.

Answer 11

Look at the second exponent rule I gave you.

Answer 12

so its going to be 8 to the power of 1 times 8 to the power of 1-log ..

Answer 13

Almost... get rid of the 1 in that second term. What you should have is the following.\[\huge 8^{1-\log_8(x+8)}=8^1\cdot 8^{-\log_8(x+8)}\]

Answer 14

Now, the second term has a negative exponent. What does that mean?

Answer 15

you divide the 2?

Answer 16

What 2?

Answer 17

u divide the 2 terms?

Answer 18

Oh... perhaps, if I am understanding you correctly. Write out what you think it is.