Exponential Equation

**Remriel** · July 1st 2012, 09:27 PM

$5^x=4^{x+1}\\\\ log(5^x)=log(4^{x+1})\\\\ (x)log5=(x+1)log4\\\\ \frac{log5}{log4}=\frac{x+1}{x}$

Where to go from here?

**Prove It** · July 1st 2012, 09:42 PM

Originally Posted by Remriel

$5^x=4^{x+1}\\\\ log(5^x)=log(4^{x+1})\\\\ (x)log5=(x+1)log4\\\\ \frac{log5}{log4}=\frac{x+1}{x}$

Where to go from here?

$\displaystyle \begin{align*} \frac{x + 1}{x} &= \frac{\log{5}}{\log{4}} \\ 1 + \frac{1}{x} &= \frac{\log{5}}{\log{4}} \\ \frac{1}{x} &= \frac{\log{5}}{\log{4}} - 1 \\ \frac{1}{x} &= \frac{\log{5} - \log{4}}{\log{4}} \\ x &= \frac{\log{4}}{\log{5} - \log{4}} \end{align*}$

**earboth** · July 1st 2012, 09:52 PM

Originally Posted by Remriel

$5^x=4^{x+1}\\\\ log(5^x)=log(4^{x+1})\\\\ (x)log5=(x+1)log4\\\\ \frac{log5}{log4}=\frac{x+1}{x}$

Where to go from here?

Your calculations are OK, but there is a simpler way to do this question:

$\displaystyle{5^x=4^{x+1}~\implies~5^x=4 \cdot 4^{x}~\implies~\left(\frac54 \right)^x=4~\implies~x = \log_{\frac54}(4)}$

With your result:

$\displaystyle{(x)log5=(x+1)log4~\implies~(x)log5=( x)log4 + log4~\implies~(x)log5-(x)log4 = log4~\implies~x=\frac{log4}{log5-log4}}$

**biffboy** · July 1st 2012, 11:48 PM

Go back to (x)log5=(x+1)log4
So (x)log5=(x)log4+log4
x(log5-log4)=log4
x=(log4)/(log5-log4}

**Remriel** · July 2nd 2012, 05:03 AM

What about this one? I can't find a way to separate: $2^{3x+1}=3^{x-2}\\\\ (3x+1)log2=(x-2)log3\\\\ \frac{3x+1}{x-2}=\frac{log3}{log2}$

**Prove It** · July 2nd 2012, 05:05 AM

Originally Posted by Remriel

What about this one? I can't find a way to separate: $2^{3x+1}=3^{x-2}\\\\ (3x+1)log2=(x-2)log3\\\\ \frac{3x+1}{x-2}=\frac{log3}{log2}$

$\displaystyle \begin{align*} \frac{3x + 1}{x - 2} &= \frac{\log{3}}{\log{2}} \\ \frac{3(x - 2) + 7}{x - 2} &= \frac{\log{3}}{\log{2}} \\ 3 + \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} \end{align*}$

Go from here...

**Remriel** · July 2nd 2012, 03:03 PM

Is this correct?

$x=\left [ \left ( \frac{log3}{log2}-3 \right )\frac{1}{7} \right ]+2$

**skeeter** · July 2nd 2012, 03:26 PM

$\frac{3x+1}{x-2} = \frac{\log{3}}{\log{2}}$

cross multiply ...

$(3x+1)\log{2} = (x-2)\log{3}$

distribute both sides ...

$3x\log{2} + \log{2} = x\log{3} - 2\log{3}$

get all x terms on the same side ...

$3x\log{2} - x\log{3} = -\log{2} - 2\log{3}$

factor out the x ...

$x(3\log{2} - \log{3}) = -(\log{2}+2\log{3})$

$x\log\left(\frac{8}{3}\right) = -\log{18}$

solve for x ...

$x = -\frac{\log{18}}{\log\left(\frac{8}{3}\right)}$

**Prove It** · July 2nd 2012, 09:00 PM

Originally Posted by Remriel

Is this correct?

$x=\left [ \left ( \frac{log3}{log2}-3 \right )\frac{1}{7} \right ]+2$

No, when you have x in the denominator, you need to reciprocate. It should be like this...

$\displaystyle \begin{align*} 3 + \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} \\ \frac{7}{x - 2} &= \frac{\log{3}}{\log{2}} - 3 \\ \frac{7}{x - 2} &= \frac{\log{3} - 3\log{2}}{\log{2}} \\ \frac{x - 2}{7} &= \frac{\log{2}}{\log{3} - 3\log{2}} \\ x - 2 &= \frac{7\log{2}}{\log{3} - 3\log{2}} \\ x &= \frac{7\log{2}}{\log{3} - 3\log{2}} + 2 \\ x &= \frac{\log{2} + 2\log{3}}{\log{3} - 3\log{2}} \end{align*}$

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