1. Log question

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

$
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?
$
log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x
$

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

$
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?
$
log_{10} \sqrt[7]{7}^{(4-3x)} = 2 log_{10} 7^x
$
base is $7$ ...

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

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

bases equal ... exponents are equal

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

solve for $x$

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

$
4-3x = 7x^2
$

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

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

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

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

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

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

$
4-3x = 7x^2
$

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

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

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

setting each factor equal to 0 ...

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