Problems from Using logarithms to simplify expressions

Solve the following equations applying logarithms:

$$3^x=17; \ 5^x\cdot7^x=\sqrt{44}$$ and $$2^{\sqrt{x}}=100$$

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Development:

In this exercise we have to add logarithms, in this case decimal, on both sides of the equality to be able to isolate $$x$$, which is in an exponent form in each cases.

$$3^x=17 \Rightarrow log3^x=log17 \Rightarrow x\cdot log3=log17 \Rightarrow$$$$$\Rightarrow x=\dfrac{log17}{log3}\simeq \dfrac{1,23}{0,477}\simeq 2,579$$$

$$5^x\cdot7^x=\sqrt{44} \Rightarrow log(5\cdot7)^x=log44^{\frac{1}{2}} \Rightarrow x\cdot(log5+log7)=\dfrac{1}{2}\cdot log44 \Rightarrow$$$$$\Rightarrow x=\dfrac{\dfrac{1}{2}\cdot log44}{log5+log7}\simeq \dfrac{0,822}{1,544} \simeq 0,532$$$

$$2^{\sqrt{x}}=100 \Rightarrow log2^{\sqrt{x}}=log100 \Rightarrow \sqrt{x}\cdot log2=2 \Rightarrow \sqrt{x}=\dfrac{2}{log2}\Rightarrow$$$$$\Rightarrow x=\Big(\dfrac{2}{log2}\Big)^2 \simeq \Big(\dfrac{2}{0,301}\Big)^2 \simeq 44,156$$$

Solution:

$$2,579; \ 0,532; \ 44,156$$

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