Solving for x

**nuckers** · May 30th 2010, 12:41 PM

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.

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

The 3x+1 is actually the exponent

**e^(i*pi)** · May 30th 2010, 12:53 PM

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.

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

The 3x+1 is actually the exponent

$(3x+1)\ln(7) = x\ln(5)$

You can distribute on the LHS as with any normal algebra

**harish21** · May 30th 2010, 12:58 PM

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.

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

The 3x+1 is actually the exponent

$7^{3x+1} = 5^x$

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

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

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

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

finish it..

**nuckers** · May 30th 2010, 01:22 PM

Originally Posted by harish21

$7^{3x+1} = 5^x$

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

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

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

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

finish it..

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

**e^(i*pi)** · May 30th 2010, 01:30 PM

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

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