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$$3^{x+1} = 3000$$

How do I solve this? I know we use logarithms but I do not remember how to solve this kind of problem. I am guessing that I need to change the problem into log form. but how?

$$\log_3{(x+1)} = 3 + \log_{10} 3$$

what do I do next?

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  • $\begingroup$ Is that supposed to be $3^{x+1}=3000$ or $3^x+1=3000$? $\endgroup$ Dec 6, 2012 at 23:38
  • $\begingroup$ $3^{x+1}=3\cdot 10^3$, divide by 3 to get $3^x=10^3$, so you got $x=\log_3(10^3)=3\log_3(10)\backsimeq 2.09590$ and this is it. $\endgroup$ Dec 6, 2012 at 23:49
  • $\begingroup$ It is 3^(x+1)=3000 $\endgroup$
    – shnisaka
    Dec 6, 2012 at 23:58
  • $\begingroup$ If you want to take base-3 logarithms, you don't get $\log_3(x+1)=3+\log_{10}3$. You get $x+1=\log_3 3000$ $=\log_3 3+\log_3 1000 = 1+\log_3 1000$. $\endgroup$ Dec 7, 2012 at 1:30

2 Answers 2

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We interpret your question as asking about $3^{x+1}=3000$. If it is about $3^x+1=3000$, rewrite as $3^x=2999$ and use the same procedure as the one below.

Take the logarithm of both sides, any base you like. I suggest base $10$ or base $e$ ($\n$), because these are easiest to find with a scientific calculator. We get $$(x+1)\log(3)=\log(3000).$$ So $$x+1=\frac{\log(3000)}{\log(3)}.$$ Calculate.

Note: We used the important fact that $\log(a^b)=b\log a$.

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  • $\begingroup$ I am talking about x+1 being as an exponent. not only x. $\endgroup$
    – shnisaka
    Dec 7, 2012 at 0:00
  • $\begingroup$ Doesn't make any difference, let $b=x+1$, $a=3$. The log of $3^{x+1}$ is $b\log a$, that is, $(x+1)\log 3$. $\endgroup$ Dec 7, 2012 at 0:02
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If you can get an equation into the form $$a^{f(x)}=b$$ for some $a,b>0$ and some function $f(x)$, then you may equivalently write $$f(x)=\log_a b,$$ or alternately, $$f(x)=\frac{\log b}{\log a}.$$ At that point, if $f(x)$ is a linear or quadratic function, you should hopefully know how to solve the resulting equation.

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