# If logx+2 5 = 10, then x is approximately I am not sure what to do with the base (x+2) but I know this: (x+2)^10 = 5 is all I know

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To do this we need some alterations done to the question.

Let `y = log_a(b)`

Then removing log will give you;

`a^y = b`

Now take log again but with b as base.

`log_b(a^y) = log_b(b)`

`ylog_b(a) = 1`

`y = 1/(log_b(a))`

Using the above we can write `log_((x+2))5` as;

`log_((x+2))5 = 1/(log_5(x+2))`

` log_((x+2))5= 10`

`1/(log_5(x+2)) = 10`

`1/10 = log_5(x+2)`

Now take the anti log;

`5^(1/10) = (x+2)`

`x = 5^(1/10)-2`

`x = -0.825`

*So the answer is x = -0.825*

**Sources:**

`log_(x+2) 5 = 10`

`(x+2)^10=5`

`x+2=e^ ln (x+2)`

So: `(x+2)^10=e^(10ln(x+2))`

`5=e^(10ln(x+2))`

Using logaritms:

`ln 5= 10ln(x+2)`

`1/10 ln 5= x+2`

`x=1/10ln5-2=-1.8390562087565899625399240666774`

`x=stackrel(-)(2).16094379124341003746007593332262`

Take the natural logarithmic equivalent:

`[In 5]/[In (x+2)]=10`

` ` Now try to isolate for x.

`In 5=10[In(x+2)]`

` ` Divide out the 10

`1/10[In 5]=In(x``+2)`

` ` Use properties of natural logarithim to to bring 1/10 as a power of In 5.

`In 5^(1/10)=In(x+2)`

` ` To solve for x take the inverse of In for both sides (e^x)

`5^(1/10)=x+2`

` ` Solve for x

`5^(1/10)-2=x`

` `