1. ## exponential equation

32 x (8*) = 3 ; give an exact solution for * (* is a power)

any help would be appreciated

2. Originally Posted by highsamurai
32 x (8*) = 3 ; give an exact solution for *

any help would be appreciated
I don't know what class this is for, but this is a good basic algebra problem. I would stop using "x" to mean multiply because in algebra you'll use "x" to mean an unknown number most of the time. You can use the "*" as a symbol for multiplication, use parentheses like 8(4) to mean 8 times 4.

Let "x" be the number that we want to find that fits the problems guidelines. It doesn't matter what it is yet, but we know it is a real number, so just call it x.

So $32*8*x=3$ meaning 32 times 8 times x equals 3. We want to rearrange this to get this in the form of x=... so let's do that. The opposite operation of multiplication is division, so if we want to "cancel" the 32, we can divide by 32 on both sides of the equation and still keep the = sign. Do the same to the 8 and you should get $x=\frac{3}{32*8}$

3. i guess i should have been more specific about the problem. the "*" is a root actually not an "x".

32 x (8*) = 3 or 8* = 3/32

x = multiply

*= power

thnx though

4. Hello, highsamurai!

Is that an exponent?

Solve for $x\!:\quad32\cdot8^x \:=\: 3$

We have: . $2^5(2^3)^x \:=\:3 \quad\Rightarrow\quad 2^{3x+5} \:=\:3$

Take logs: . $\ln\left(2^{3x+5}\right) \:=\:\ln(3) \quad\Rightarrow\quad(3x+5)\ln(2) \:=\:\ln(3)$

. . $3x+5 \:=\:\frac{\ln(3)}{\ln(2)} \quad\Rightarrow\quad 3x \:=\:\frac{\ln(3)}{\ln(2)} - 5 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{1}{3}\left(\frac{\ln(3)}{\ln(2)} - 5\right)}$

5. thnx man that helped a lot.

I also have another one if you dont mind. I just dont know what to do next.

log10 (x - 5) + log10 (3x + 6) = log10 (x+4)

log10 (x - 5)(3x + 6) = log10 (x + 4)

(x - 5)(3x + 6) = (x + 4)

3x^2 - 9x - 30 = x + 4

3x^2 - 9x - 34 = x

(the ^2 is squared)
Im lost after that, i dont know how to factor that.

(the "10's" should be lowered)