1. ## Log question

The problem:
$\displaystyle \sqrt[7]{7}^{(4-3x)} = 7^{x^2}$

$\displaystyle x = -1, \frac{4}{7}$

How do you do this problem?

It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
$\displaystyle log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x$

2. Originally Posted by desiderius1
The problem:
$\displaystyle \sqrt[7]{7}^{(4-3x)} = 7^{x^2}$

$\displaystyle x = -1, \frac{4}{7}$

How do you do this problem?

It says the first thing I need to do is to get both sides to have the same base. Wouldn't the base be 10?
$\displaystyle log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x$
base is $\displaystyle 7$ ...

note that $\displaystyle \sqrt[7]{7} = 7^{\frac{1}{7}}$

$\displaystyle \left(7^{\frac{1}{7}}\right)^{4-3x} = 7^{x^2}$

bases equal ... exponents are equal

$\displaystyle \frac{4-3x}{7} = x^2$

solve for $\displaystyle x$

3. $\displaystyle \frac{4-3x}{7} = x^2$

$\displaystyle 4-3x = 7x^2$

$\displaystyle 7x^2+3x-4=0$

$\displaystyle (7x+1)(x-\frac{4}{7})$

I get -4/7 and not -4, is this right?

$\displaystyle 7x^2-4x+x-\frac{4}{7}$

$\displaystyle 7x^2-3x-\frac{4}{7}$

4. Originally Posted by desiderius1
$\displaystyle \frac{4-3x}{7} = x^2$

$\displaystyle 4-3x = 7x^2$

$\displaystyle 7x^2+3x-4=0$

$\displaystyle (7x+1)(x-\frac{4}{7})$ ???
$\displaystyle 7x^2 + 3x - 4 = 0$

$\displaystyle (7x - 4)(x + 1) = 0$

setting each factor equal to 0 ...

$\displaystyle x = \frac{4}{7}$ , $\displaystyle x = -1$