# Exponential equation - logarithmisation

is the transformation of this equation: $$9^x + 6^x = 2× 4^x$$ into this: $$\log_2 (9^x) + \log_2 (6^x)=\log_2 (2×4^x)$$ correct? I want to know because I really want to solve this equation.

• I am afraid not, problem is the left hand side. Logs do not distribute that way – imranfat Feb 13 '18 at 17:15
• It is not correct. $$\log_2 (9^x) + \log_2(6^x) = \log_x(9^x \cdot 6^x).$$ So $\log_2 (9^x) + \log_2(6^x) = \log_x(2\times 4^x)$ is equivalent to $9^x\times6^x = 2\times 4^x.$ $${}$$ – Michael Hardy Feb 13 '18 at 17:15
• So should it be this instead? $$\log_2 (9^x + 6^x)=\log_2 (2×4^x)$$ – Robert874 Feb 13 '18 at 17:17
• Robert874: Please explain how the "answer" you saw fit to accept is addressing the question "is the transformation of this equation: $9^x + 6^x = 2× 4^x$ into this: $\log_2 (9^x) + \log_2 (6^x)=\log_2 (2×4^x)$ correct? I want to know because I really want to solve this equation." – Did Feb 13 '18 at 23:03

HINT: write your equation in the form $$\left(\frac{3}{2}\right)^{2x}+\left(\frac{3}{2}\right)^x=2$$

• (-1). This hint absolutely does not address the question asked, which is whether the asker applied the logarithm correctly. A good answer would correct the error and point out the mistake. – user296602 Feb 13 '18 at 21:23
• i will note this in the future! – Dr. Sonnhard Graubner Feb 15 '18 at 22:46

No. One property of logarithms is that $\log(a) + \log(b) =\log(ab).$

So $\log(a + b)$ may not be reduced as you have done.

It's $f(x)=0$, where $$f(x)=\left(\frac{3}{2}\right)^{2x}+\left(\frac{3}{2}\right)^{x}-2.$$ We see that $f$ increases, which says that our equation has one root maximum.

But, $0$ is a root and we are done!

Your reasoning is wrong because $\log(a+b)$ is not always equal to $\log{a}+\log{b}.$