Question

Solve -2log8(x + 1) = -8

Answer 1

isolate the logrithm, so divide both sides by -2,

then make both sides of the equation exponents to the number 8 to cancel the logarithm

Answer 2

can you show me step by step ?please

Answer 3

-2log8(x + 1) = -8
=
log_8(x+1) = -8/-2
=
\[8^{\log_8(x+1)}=8^{-8/-2} \]

remember the rule:

\[y^{\log_y(x)} = x\]
another example of this rule:
\[3^{\log_3(x+1)} = x+1\]
etc

Answer 4

so you end up with

x+1 = 8^(8/2)

Answer 5

now you can use simple algebra to solve

Answer 6

just remember that rule, and the fact that as long as you do the same operation to both sides of the equation the equation maintains its equality

Answer 7

thank you :)

Answer 8

no problem any other questions feel free to ask

Answer 9

just one more
Write 2log35 + log32 as a single logarithm ?

Answer 10

use the rule

\[xlog(3) = \log(3^{x})\]
an example of this rule:
\[3\log(x+1) = \log((x+1)^{3})\]

and the rules (remember you can only do this with logarithms that are the same)

\[\log(x) + \log(y) = \log(xy)\]
and
\[\log(x) - \log(y) = \log(\frac{x}{y})\]

you cannot do this though

\[\log_8(x) + \log(y) \neq \log(xy)\]

Answer 11

do you follow?
an example of the last two rules is

\[\log(s) - \log(y) + \log(3x) - \log(m)= \log(\frac{3xs}{ym})\]

Answer 12

2log(base3)(7) is this right

Answer 13

I think it may just be log(37)

\[\log(x) = \log_{10}(x)\]

Answer 14

people are lazy so we just call log(x) base 10

Answer 15

unless the the 3 is in the subscript then it is

\[\log_3(x)\]

Answer 16

I'm assuming

\[ 2\log_{10}35 + \log_{10}(32)\]

Answer 17

so

the first step
\[\log(35^{2}) + \log(32)\]

Answer 18

2log3^5 + log3^2 this the question sorry

Answer 19

\[2\log(3^{5}) + \log(3^{2})\]
so this is the question

Answer 20

?

Answer 21

yes

Answer 22

here is the first step just use the rules I showed you I will tell you if you get the right answer
\[\log(3^{5(2)}) + \log(3^{2})\]