Find 'x' satisfying equation $4^{\log_{10} {x+1}} - 6^{\log_{10} x} - 2.3^{\log_{10} {x^2 +2}}$ = 0

Question

The question is from logs

Find 'x' satisfying equation $4^{\log_{10} {x+1}} - 6^{\log_{10} x} - 2.3^{\log_{10} {x^2 +2}}$ = 0

I've tried to solve it. I was trying to convert base 10 of logs to respective numbers. like 4 for first one 6 for second one and 2.3 for third one. I thought that I'll be able to move those logs from powers. But I'm not able to figure how I convert them .

Please explain I can solve it. And Please solve it like a class 11th student.

Answer 1

The above expression is reducible to:

$$4\times 2^{2\log(x)} - 6^{\log(x)} - 18 \times 3^{2 \log(x)} = 0$$

Now let $ 2^{\log(x)} = a$ and $3^{\log(x)} = b$.

The above expression then becomes:

$$4a^2-ab-18b^2 = 0$$

Now factorize and solve for $x$!

Answer 2

Hint: Note that

$$ \text{log}_{10}(x) = \frac{\text{log}_{a}(x)}{\text{log}_{a}(10)} $$

for any $a>0$. You could use this fact to your advantage by doing:

$$ a^{\text{log}_{10}(x)} = a^\frac{{\text{log}_{a}(x)}}{\text{log}_a(10)} = \left( a^{\text{log}_{a}(x)}\right)^{\frac{1}{\text{log}_a(10)}} = x^{\frac{1}{\text{log}_a(10)}} $$

Answer 3

Hint.

Calling

$$ a=2^{\log_{10}x}\\ b=3^{\log_{10}x} $$

we have

$$ 2^2a^2-a b-2\cdot 3^2b^2=(4a-9b)(a+2b) = 0 $$