1. ## Solving for x

Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

$\displaystyle 7 ^3x+1=5^x$

The 3x+1 is actually the exponent

2. Originally Posted by nuckers
Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

$\displaystyle 7 ^3x+1=5^x$

The 3x+1 is actually the exponent
$\displaystyle (3x+1)\ln(7) = x\ln(5)$

You can distribute on the LHS as with any normal algebra

3. Originally Posted by nuckers
Can anyone help me out on this equation, i am doing exponents and logarithms, and can't seem to understand this, i get that you are supposed to make the base the same and then solve for x using the exponents, but how do you make the base the same in this function.

$\displaystyle 7 ^3x+1=5^x$

The 3x+1 is actually the exponent
$\displaystyle 7^{3x+1} = 5^x$

$\displaystyle log(7^{3x+1}) = log(5^x)$

$\displaystyle (3x+1) \mbox{log(7)} = x \mbox{log(5)}$

$\displaystyle \frac{3x+1}{x}= \frac{\mbox{log(5)}}{\mbox{log(7)}}$

$\displaystyle 3+ \frac{1}{x} = \frac{\mbox{log(5)}}{\mbox{log(7)}}$

finish it..

4. Originally Posted by harish21
$\displaystyle 7^{3x+1} = 5^x$

$\displaystyle log(7^{3x+1}) = log(5^x)$

$\displaystyle (3x+1) \mbox{log(7)} = x \mbox{log(5)}$

$\displaystyle \frac{3x+1}{x}= \frac{\mbox{log(5)}}{\mbox{log(7)}}$

$\displaystyle 3+ \frac{1}{x} = \frac{\mbox{log(5)}}{\mbox{log(7)}}$

finish it..
So i get -.46021234 for my final answer, does that sound right?

5. Originally Posted by nuckers
So i get -.46021234 for my final answer, does that sound right?
That's correct ^_^

You can test it by putting it into your original equation. If the two sides are equal than the answer is fine