3log8(2x+7)+8=10

## solve for x

# 3 Answers

With any question in this form, we always first completely isolate the 'log' part of the function, and then solve from there. To get the log_{8}(2x+7) alone, we solve for it just as we would solve for any other variable.

**3log _{8}(2x+7)+8=10**

-8 -8

**3log _{8}(2x+7) = 2**

/3 /3

**log _{8}(2x+7) =2/3 **

Here we need to remove the log term. We know that log_{b}(x) = n is the same thing as saying b^{n }= x. So, we raise 8 to the power of both sides. This gets rid of the log_{8 }term. So:

**2x + 7 = 8 ^{2/3}**

**2x+7 = 4** and from here we can solve like any other equation

-7 -7

**2x = -3**

/2 /2

**x = -3/2**

The generic form of a log and exponent form can be written:

**log _{b}x = n** which also mean that

**b**

^{n}= xStart by getting the equation into the format of log_{b}x = n

1.) Subtract 8 from both sides: 3log_{8}(2x+7) = 2

2.) Divide both sides by 3: log_{8}(2x+7) = 2/3

3.) Use the relationship between logs and exponents to rewrite the equation without the log

b^{n} = x; where b = 8, x = (2x + 7), and n = 2/3

8^{(2/3)} = 2x + 7

4.) Solve for x

4 = 2x + 7

-3 = 2x

-3/2 = x

So your solution is x = -3/2

Hope this will help you with this problem type in the future!

Isolate log8(2x+7) first,

log8(2x+7) = (10-8)/3 = 2/3

Change to exponential form,

2x+7 = 8^(2/3) = 4

Solve for x,

x = (4-7)/2 = -3/2 <==Answer