Logarithm — JEE Main practice

Factor the expressions:

$$6𝑥^2+28𝑥+30=2(𝑥+3)(3𝑥+5),$$ $$𝑥^2+6𝑥+9=(𝑥+3)^2.$$

Using logarithm properties,

$$\log_{(𝑥+3)}2+\log_{(𝑥+3)}(𝑥+3)+\log_{(𝑥+3)}(3𝑥+5)+2\log_{(6𝑥+10)}(𝑥+3)=5.$$

Since

$$\log_{(𝑥+3)}2+1=\log_{(𝑥+3)}(6𝑥+10),$$

the equation becomes

$$\log_{(𝑥+3)}(6𝑥+10)+4\log_{(6𝑥+10)}(𝑥+3)=4.$$

Let

$$t=\log_{(𝑥+3)}(6𝑥+10).$$

Then

$$\log_{(6𝑥+10)}(𝑥+3)=\frac1t,$$

so

$$t+\frac4t=4 \Rightarrow t^2-4t+4=0 \Rightarrow (t-2)^2=0.$$

Hence,

$$t=2.$$

Checking the original equation gives the valid solutions

$$𝑥=1,\,-1.$$

Therefore,

$$1+(-1)=0.$$

First, evaluate the infinite geometric series.

For

$$\alpha=\frac14+\frac18+\frac1{16}+\cdots,$$

the first term is

$$a=\frac14,\qquad r=\frac12.$$

Hence,

$$\alpha=\frac{a}{1-r} =\frac{\frac14}{1-\frac12} =\frac12.$$

Similarly,

$$\beta=\frac13+\frac19+\frac1{27}+\cdots$$

has

$$a=\frac13,\qquad r=\frac13,$$

so

$$\beta=\frac{\frac13}{1-\frac13} =\frac12.$$

Therefore, the given expression becomes

$$\left(\frac15\right)^{\log_{\sqrt5}\left(\frac12\right)} +\left(\frac1{25}\right)^{\log_5\left(\frac12\right)}.$$

Using the change of base formula,

$$\log_{\sqrt5}\left(\frac12\right) =\frac{\log_5\left(\frac12\right)}{\log_5(\sqrt5)} =2\log_5\left(\frac12\right).$$

Hence,

$$\left(\frac15\right)^{\log_{\sqrt5}\left(\frac12\right)} =5^{-2\log_5(1/2)} =5^{\log_5 4} =4.$$

Also,

$$\left(\frac1{25}\right)^{\log_5(1/2)} =5^{-2\log_5(1/2)} =5^{\log_5 4} =4.$$

Therefore,

$$4+4=8.$$

Hence, the correct option is C.

We begin with the equation

$$e^{5(\log_e 𝑥)^2+3}=𝑥^8,\qquad 𝑥>0.$$

Since

$$𝑥^8=e^{8\log_e 𝑥},$$

equate the exponents:

$$5(\log_e 𝑥)^2+3=8\log_e 𝑥.$$

Let

$$𝑡=\log_e 𝑥.$$

Then

$$5𝑡^2-8𝑡+3=0.$$

Factorizing,

$$(5𝑡-3)(𝑡-1)=0.$$

Hence,

$$𝑡=1 \quad\text{or}\quad 𝑡=\frac35.$$

Therefore,

$$𝑥=e \quad\text{or}\quad 𝑥=e^{3/5}.$$

The product of all solutions is

$$e\cdot e^{3/5} =e^{1+\frac35} =e^{8/5}.$$

Hence, the correct option is D.