# Solve: 4^3-x = 8^x (can you explain how to do it using natural log if able to use it in this equation) thank you

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Given problem is `4^(3-x)=8^x`

Now taking log of both sides we get

`ln(4^(3-x))=ln(8^x)` . Because `ln(m^n)=n lnm`

So, `(3-x)ln4=xln8`

or, `(3-x)ln(2^2)=xln(2^3)`

or, `2(3-x)ln2=3xln2`

or, `2(3-x)ln2-3xln2=0`

or, `{2(3-x)-3x}ln2=0`

As `ln2!=0` . So, {2(3-x)-3x}=0

or, `2.3-2x-3x=0`

or, `6-5x=0`

or, `5x=6`

or, `x=6/5` . Answer.

We can write the 4 and 8 as an exponential expression with base 2.

So, we will have:

`2^(2(3-x)) = 2^(3x)`

`2^(6 - 2x) = 2^(3x)`

Since the bases are the same, we can just equate the exponents.

`6 - 2x = 3x`

Add both sides by 2x.

`6 = 5x`

Divide both sides by 5.

Therefore, **x = 6/5**.

`4^3 -x=8^x` we study the function:

`f(x)= 8^x +x - 4^3` the first derivative is `f'(x)= 8^(x)lnx +1` the second: `f''(x)= 8^(x)ln^2x +8^(x)/x`

Now if `f'(x_0)=0` then: `8^(x_0)= -1/ln(x_0)` (1)

subistutes in the second derivative:

`f''(x_0)= -ln(x_0) -1/(x_0 ln(x_0))` `=-(x_0 ln^2(x_0)+1)/(ln(x_0))` (2)

since `x_0>0` the ratio is positive, if `0< x_0 <=1` (minimum point)

and negative if `x_0 >1` (Maximus point)

Now `f(x_0)= 8^(x_0) +x_0 -4^3` `=(8^(x_0)ln(x_0)+x_0 ln(x_0))/(ln(x_0))` `-4^3=`

`=(-1+ln(x_0)(x_0 -4^3))/(ln(x_0))` (3)

This last realtion show that if has a minimum is less than zero.

Instead if maximus to have solution needs `x_0 >= 4^3` but it's clear that this one doesn't accomplish relation (1) so if exists a point `x_0` is to be a minimum point.