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

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.

• Maybe you could add some parentheses to make it clearer, for example, if the first term is $4^{\log_{10}(x+1)}$ or $4^{(\log_{10}x)+1}$? Anyway, in the latter case you can do $$4^{(\log_{10}x)+1}=4\cdot 4^{\log_{10}x}=4\cdot\left(10^{\log_{10}4}\right)^{\log_{10}x}=4\cdot\left(10^{\log_{10}x}\right)^{\log_{10}4}=4x^{\log_{10}4}$$ but I'm not sure how helpful this is. – Luiz Cordeiro May 1 '19 at 19:53

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 $$x$$!

• Please explain, how did you solved it to $18 × 3^{2\log(x)}$ – Piyush Raj May 1 '19 at 20:15
• $2\times 3^2 \times 3^{\log(x^2)} = 2\times3^{\log(x^2) +2}$ – Vizag May 1 '19 at 20:19
• Okay I got it. Maybe you didn't noticed it. That isn't multiply in between 2.3 ,its decimal. – Piyush Raj May 1 '19 at 20:21
• That's highly unlikely that it's a decimal. Solve it assuming $\times$ and see if you get the answer given? – Vizag May 1 '19 at 20:26
• :( Its really hard to factorize it. – Piyush Raj May 1 '19 at 20:36

Hint.

Calling

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

we have

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

• Maybe you didn't saw it properly its isn't 2⋅3, its 2.3. Decimal is in between. – Piyush Raj May 1 '19 at 20:03

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:

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