How to find the value of x in:
$$10=8.4\log(0.3x+1)$$
so far I got :
$$10=\log(0.3x+1)^{8.4}$$
$$10^{10}=(0.3x+1)^{8.4}$$
What should I do next?
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Raise both sides to the power $1/8.4$. – pshmath0 Mar 09 '14 at 07:22
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Presumably you $\log$ is base $10$? – pshmath0 Mar 09 '14 at 07:23
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Here is a hint: Start with $$10=8.4\log(0.3x+1).$$ Divide both sides by $8.4$ too get $$\frac{10}{8.4}=\log(0.3x+1).$$ Then use the fact that $$10^{log(Y)}=Y$$ for $Y>0$. – Baby Dragon Mar 09 '14 at 07:25
2 Answers
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${10 \over 8.4} = \log (0.3 x +1 )$.
$10^{10 \over 8.4} = 0.3 x +1$.
${ 10^{10 \over 8.4} -1 \over 0.3} = x$.
copper.hat
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Our equation: $$10=8.4\log(0.3x+1)$$ Divide $8.4$ on both sides $$\dfrac{10}{8.4}=\log(0.3x+1)$$ $$\dfrac{100}{84}=\log(0.3x+1)$$ $$\dfrac{25}{22}=\log(0.3x+1)$$ Remember that if $\log_a(b)=x$, then $a^x=b$. $$\dfrac{25}{22}=\log_{10}(0.3x+1)$$ $$10^{25/22}=0.3x+1$$ $$\sqrt[22]{10^{25}}=0.3x+1$$ $$0.3x=\sqrt[22]{10^{25}}-1$$ $$\dfrac{3}{10}x=\sqrt[22]{10^{25}}-1$$ $$x=\dfrac{\sqrt[22]{10^{25}}-1}{\left(\dfrac{3}{10}\right)}$$ $$\displaystyle \boxed{x=\dfrac{10\sqrt[22]{10^{25}}-1}{3}}$$
Anonymous Computer
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