Solving for x

Jan 2010
60
0
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
 

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
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
 
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Feb 2010
1,036
386
Dirty South
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..
 
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Jan 2010
60
0
\(\displaystyle 7^{3x+1} = 5^x\)

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

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

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

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

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

e^(i*pi)

MHF Hall of Honor
Feb 2009
3,053
1,333
West Midlands, England
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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