Is it possible to use exponents as a form of addition? For example in 51.x=65^{1.x} = 6 where xx is the numbers after the decimal place. If so, what would be the equation to find xx?

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Not clear on your notation. What is the exponent? If, say, x=2x=2, what is the left hand of your expression?

– lulu

2 days ago

1

Yes. If I get your notation right, you can take log\log from both sides and rewrite 1.x1.x as 1+x101+\frac{x}{10}, which comes down from the exponent. Then solve linearly for xx.

– O. Von Seckendorff

2 days ago

1

For your example, x≈0.1133x \approx 0.1133.

– O. Von Seckendorff

2 days ago

Thanks all! Especially O. Von Seckendorff. Lulu: for clarification I meant (as an example) 5^1.1133

– Indigo2003

2 days ago

O. Von Seckendorff what do you mean log (I know this means logarithm) both sides? Wouldn’t I need an exponent and base etc?

– Indigo2003

2 days ago

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1 Answer

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So I think you are asking how do you solve ba+x=kb^{a + x} = k for xx if b,a,kb,a,k are constants; for example how would you solve 51+x=65^{1 + x} = 6. Is that right.

Like so:

51+x=65^{1 + x} = 6

log551+x=log56\log_5 5^{1+x}= \log_5 6

1+x=log561 +x = \log_5 6

x=log56−1x = \log_5 6 – 1

x=ln6ln5−1x = \frac{\ln 6}{\ln 5} – 1

x=1.113…−1=0.113…x = 1.113…-1=0.113…

In general ba+x=kb^{a + x} = k

if x=logbk−ax = \log_b k – a

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To figure out what logbk\log_b k is … logbk=logxklogxb=log10klog10b=logeklogeb=lnklnb\log_b k = \frac{\log_x k}{\log_x b} = \frac{\log_{10} k}{\log_{10} b}= \frac{\log_e k}{\log_e b}= \frac{\ln k}{\ln b}

how did you get from x = log5 6 – 1 to x = in… and why are we using in?

– Indigo2003

2 days ago

ln\ln = “natural log” =loge \log_e; e = 2.717……. logbk=logxk/logxb\log_b k = \log_x k/\log_x b for any x whatsoever. log56=log106/log105=loge6/loge5=log29.7536/log29.76535 \log_5 6 = \log_{10} 6/\log_{10} 5 = \log_e 6/\log_e 5 = \log_{29.753} 6/\log_{29.7653} 5. Chances are you do not have a log5\log_5 button on your calculator. Doesn’t matter as you probably have a ln\ln button. And you can always solve logbk=lnklnb\log_b k = \frac{\ln k}{\ln b} you can also solve logbk=log10klog10b\log_b k = \frac{\log_{10} k}{\log_{10} b}. But log10\log_{10} is what *physicists* use. Real *mathematicians* use ln=loge\ln = \log_e. 🙂

– fleablood

2 days ago

If bk=Nb^k = N then logbN=k\log_b N =k. That’s a definition. By definition xlogxb=bx^{\log_x b} = b. So if bk=Nb^k = N then xk∗logxb=Nx^{k*\log_x b} = N so logxxklogxb=logxN\log_x x^{k\log_x b} = \log_x N so klogxb=logxNk \log_x b = \log_x N so k=logxNlogxbk = \frac{\log_x N}{\log_x b} so k=logbN=logxNlogxbk = \log_b N =\frac{\log_x N}{\log_x b}. So log56=log105log106=ln5ln6\log_5 6 = \frac{\log_{10}5}{\log_{10}6}=\frac{\ln 5}{\ln 6}.

– fleablood

2 days ago

Thanks! I think my brain just broke though… Also, why do we use e? And I know it is 2.718281828459.

– Indigo2003

2 days ago

And I thought (Phi^n – phi^n)/sqrt5 was complicated…. Can someone tell me how to use division bars and superscript please?

– Indigo2003

2 days ago