# I am trying to solve the following question(s): Solve: 4 ^ 8x-1 = 8 3 ^ 2x-5= 1/27^x This question involves logs. I am doing this math course via distance learning on computer and have no...

I am trying to solve the following question(s):

Solve:

4 ^ 8x-1 = 8

3 ^ 2x-5= 1/27^x

This question involves logs. I am doing this math course via distance learning on computer and have no idea how to answer the question. I have looked on many websites and 3 of my text books and I just don't get it. Thank you.

*print*Print*list*Cite

### 1 Answer

(1) Solve `4^(8x-1)=8`

(a) you can write each side as an exponent with the same base:

`4^(8x-1)=(2^2)^(8x-1)=2^(2(8x-1))=2^(16x-2)` and `8=2^3` . Thus:

`2^(16x-2)=2^3` Since the bases are equal the exponents are equal:

`16x-2=3==>16x=5==>x=5/16`

Checking we see that `4^(8*5/16-1)=4^(5/2-1)=4^(3/2)=(sqrt(4))^3=2^3=8` as required.

(b) Using logarithms:

`4^(8x-1)=8` Take a logarithm of each side: I will use the natural logarithm (with base `e` ) written as `lnx`

**Why use `lnx` ? You can use any logarithm -- most calculators only have keys for the natural log and the common log, so I use one of these two. **

`ln(4^(8x-1))=ln8` A property of logs is `log_c a^x=xlog_c a` so:

`(8x-1)ln4=ln8`

`8x-1=(ln8)/(ln4)`

`8x-1=1.5` from calculator

`8x=2.5==>x=2.5/8=5/16`

-------------------------------------------------------------

**The solution to `4^(8x-1)=8` is `x=5/16` **

-------------------------------------------------------------

(2) Solve `3^(2x-5)=1/(27^x)`

(a) Again we can rewrite each side with the same base. (This is not always possible)

`1/(27^x)=27^(-x)` using the property of exponents that `a^(-m)=1/a^m`

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

So `3^(2x-5)=3^(-3x)` Same bases so the exponents are equal:

`2x-5=-3x`

`5x=5==>x=1`

Checking we have `3^(2-5)=3^-3=1/27` as required.

(b) Using logarithms:

`ln(3^(2x-5))=ln(1/(27^x))`

The left hand side becomes `(2x-5)ln3`

For the right hand side, we need a property of logarithms: `log(A/B)=logA-logB` , so:

`ln(1/(27^x))=ln1-ln27^x` Now `ln1=0` (true for any base) so we have:

`(2x-5)ln3=-xln27`

`(2x-5)/x=-(ln27)/(ln3)`

`(2x-5)/x=-3` from calculator

`2x-5=-3x==>5x=5==>x=1` as before.

--------------------------------------------------------------

**The solution to `3^(2x-5)=1/(27^x)` is x=1.**

-------------------------------------------------------------

**Sources:**