Tricky logarithmic question

Question

I have been asked to find $x$ satisfying the equation $$4^{(\log x)+1}-6^{\log x}-2\times3^{(\log x^2)+2}=0.$$

I am totally clueless how to solve this, so any hint will be appreciated.

Answer 1

I get $x = \frac{1}{100}$ for your latest version (posted around midnight CST), and substitution into your equation shows that $\frac{1}{100}$ is a solution. My solution in great detail is given below. However, I suspect there is probably a quicker way to solve this equation than going through all the algebraic contortions I went through.

(ADDED ABOUT AN HOUR LATER) Glancing over what I did suggested to me a quicker way, which I'll outline in this paragraph and leave my original solution alone. Letting $u = \log_{10}x,$ the equation becomes $4^{u+1} - 6^u - 18\cdot 9^u = 0.$ Some rewriting leads to $4\cdot \left(2^u\right)^2 - 2^u \cdot 3^u - 18 \cdot \left(3^u\right)^2 = 0,$ and this factors as $\left(2^u + 2\cdot 3^u\right)\left(4\cdot 2^u - 9 \cdot 3^u\right) = 0,$ which is now easy to solve for $u$ (same way I do near the end below), and now use $x = 10^{u}.$

In what follows, keep in mind that $x > 0$ holds, since otherwise two of the logarithms in the original equation are not defined. This will be used in at least two places, one being for the identity $\log_{10} x^2 = 2\log_{10}x$ and the other near the end when asserting that $x^{\log_{10} \frac{2}{3}} = -2$ has no solution.

$$ 4^{(\log_{10} x) + 1} \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 2 \times 3^{(\log_{10} x^2) + 2} \;\; = \;\; 0 $$

$$ 4^{\log_{10} x}\times 4^1 \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 2 \times 3^{(\log_{10} x^2)} \times 3^2 \;\; = \;\; 0 $$

$$ 4 \times 4^{\log_{10} x} \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 18 \times 3^{(\log_{10} x^2)} \;\; = \;\; 0 $$

$$ 4 \times 4^{\log_{10} x} \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 18 \times 3^{(2\log_{10} x)} \;\; = \;\; 0 $$

$$ 4 \times 4^{\log_{10} x} \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 18 \times \left(3^2\right)^{\log_{10} x} \;\; = \;\; 0 $$

$$ 4 \times 4^{\log_{10} x} \;\; - \;\; 6^{\log_{10} x} \;\; - \;\; 18 \times 9^{\log_{10} x} \;\; = \;\; 0 $$

We now make use of the identity

$$a^{\log_{10}c} = c^{\log_{10}a},$$

which I mentioned in a comment and which can be justified as follows: $a^{\log_{10}c} = a^{\frac{\log_a c}{\log_a {10}}}$ (in the exponent, change base-$10$ logarithm to base-$a$ logarithms) AND $ a^{\frac{\log_a c}{\log_a {10}}} = \left( a^{\log_a c}\right)^{\frac{1}{\log_a {10}}}$ (use the identity $a^{VW} = \left(a^V\right)^W)$ AND $\left( a^{\log_a c}\right)^{\frac{1}{\log_a {10}}} = c^{\frac{1}{\log_a {10}}}$ (use the identity $a^{\log_a c} = c)$ AND $c^{\frac{1}{\log_a {10}}} = c^{\log_{10} a}$ (use the identity $\log_b a = \frac{1}{\log_a b}).$

Applying this identity gives

$$ 4 x^{\log_{10} 4} \;\; - \;\; x^{\log_{10} 6} \;\; - \;\; 18 x^{\log_{10} 9} \;\; = \;\; 0 $$

Using $\; \log_{10} 4 = \log_{10} 2^2 = 2\log_{10} 2 \; $ AND $\; \log_{10} 6 = \log_{10} {(2 \cdot 3)} = \log_{10} 2 + \log_{10} 3 \;$ AND $\; \log_{10} 9 = \log_{10} {(3^2)} = 2 \log_{10} 3, \;$ we get

$$ 4 x^{2\log_{10} 2} \;\; - \;\; x^{\log_{10} 2 + \log_{10} 3} \;\; - \;\; 18 x^{2\log_{10} 3} \;\; = \;\; 0 $$

$$ 4 \left(x^{\log_{10} 2}\right)^2 \;\; - \;\; x^{\log_{10} 2} \cdot x^{\log_{10} 3} \;\; - \;\; 18 \left(x^{\log_{10} 3}\right)^2 \;\; = \;\; 0 $$

Letting $\;A = x^{\log_{10} 2}\;$ and $\;B = x^{\log_{10} 3},\;$ this last equation becomes

$$4A^2 \; - \; AB \; - \; 18B^2 $$

The left side can be factored:

$$(A + 2B)(4A - 9B) \; = \; 0$$

Therefore, $\; A = -2B\;$ and $\;4A = 9B,\;$ which gives

$$ x^{\log_{10} 2} = -2x^{\log_{10} 3} \;\;\;\; \text{and} \;\;\;\; 4x^{\log_{10} 2} = 9x^{\log_{10} 3} $$

Dividing both sides of each equation by $x^{\log_{10} 3}$ gives

$$ \frac{x^{\log_{10} 2}}{x^{\log_{10} 3}} = -2 \;\;\;\; \text{and} \;\;\;\; \frac{4x^{\log_{10} 2}}{x^{\log_{10} 3}} = 9 $$

$$ x^{\log_{10} 2 - \log_{10} 3} = -2 \;\;\;\; \text{and} \;\;\;\; 4x^{\log_{10} 2 - \log_{10} 3} = 9 $$

$$ x^{\log_{10} \frac{2}{3}} = -2 \;\;\;\; \text{and} \;\;\;\; 4x^{\log_{10} \frac{2}{3}} = 9 $$

Since $x > 0$ (see very beginning above), the left equation has no solution. To solve the right equation, we first divide both sides by 4 and then we raise both sides to the $\frac{1}{\log_{10} \frac{2}{3}}$ power:

$$x^{\log_{10} \frac{2}{3}} = \frac{9}{4} $$

$$ x = \left(\frac{9}{4}\right)^{ \frac{1}{\log_{10} \frac{2}{3}} } $$

To see if this simplifies, let's take base-$10$ logarithms of both sides and see what happens:

$$ \log_{10} x \; = \; \log_{10} \left[ \left(\frac{9}{4}\right)^{ \frac{1}{\log_{10} \frac{2}{3}} } \right] $$

$$ \log_{10} x \; = \; \frac{1}{\log_{10} \frac{2}{3}} \cdot \log_{10} \left(\frac{9}{4}\right) \; = \; \frac{ \log_{10}\frac{9}{4} } { \log_{10} \frac{2}{3} }$$

$$ \log_{10} x \; = \; \frac{ \log_{10} \left( \frac{2}{3} \right)^{-2} } { \log_{10} \frac{2}{3} } \; = \; \frac{ -2\log_{10} \frac{2}{3} } { \log_{10} \frac{2}{3} } \; = \; -2$$

Therefore, we have

$$x \; = \; 10^{-2} \; = \; \frac{1}{100} $$

As a safety check, let's see if this satisfies the original equation

$$ 4^{(\log x) + 1} \;\; - \;\; 6^{\log x} \;\; - \;\; 2 \times 3^{(2\log x) + 2} \;\; = \;\; 0 $$

Substituting $\log_{10} x = -2$ into the left side above gives

$$ 4^{-2 + 1} \;\; - \;\; 6^{-2} \;\; - \;\; 2 \times 3^{2(-2) + 2} $$

$$ = \;\; 4^{-1} \;\; - \;\; 6^{-2} \;\; - \;\; 2 \times 3^{-2} $$

$$ = \;\; \frac{1}{4} \;\; - \;\; \frac{1}{36} - \;\; \frac{2}{9} $$

$$ = \;\; \frac{9}{36} \;\; - \;\; \frac{1}{36} - \;\; \frac{8}{36} $$

$$ = \;\; 0 $$

Answer 2

Unfortunately, the current equation has no real solutions for $x$.

Answer 3

Since $\log$ can have different meanings, let $\log=\log_a$ for some $a>1$ (most common values for $a$ are $\mathrm{e},10$ and $2$).

\begin{align} 4^{\log_a(x) +1}-6^{\log_a(x)}-2\cdot3^{\log_a(x^2)+2}&=0 \tag{1}\label{1} ,\\ 4^1\cdot (2\cdot2)^{\log_a(x)} -(2\cdot3)^{\log_a(x)} -2\cdot3^2\cdot3^{2\log_a(x)}&=0 ,\\ 4\cdot 2^{\log_a(x)}\cdot 2^{\log_a(x)} -2^{\log_a(x)}\cdot3^{\log_a(x)} -18\cdot3^{\log_a(x)}\cdot3^{\log_a(x)}&=0 \tag{2}\label{2} . \end{align}

Now, let's divide \eqref{2} by $2^{\log_a(x)}\cdot3^{\log_a(x)}$:

\begin{align} 4\cdot\frac{2^{\log_a(x)}}{3^{\log_a(x)}} -1 -18\cdot \frac{3^{\log_a(x)}}{2^{\log_a(x)}}&=0 ,\\ 4\cdot\left(\frac{2}{3}\right)^{\log_a(x)} -1 -18\cdot\frac{1}{\left(\frac23\right)^{\log_a(x)}} &=0 . \end{align}

Let $\left(\frac{2}{3}\right)^{\log_a(x)}=t$. Then we have

\begin{align} 4t-1 -18\cdot\frac{1}{t} &=0 ,\\ 4\,t^2-t-18&=0 \tag{3}\label{3} . \end{align}

Quadratic equation \eqref{3} has two solutions, \begin{align} t_1&=\tfrac94 ,\\ t_2&=-2 , \end{align}

but we know that $t$ must be positive, so the only option left is

\begin{align} \left(\frac{2}{3}\right)^{\log_a(x)} &=\frac94 =\left(\frac23\right)^{-2} ,\\ \log_a(x)&=-2 ,\\ x&=a^{-2} . \end{align}

So if $\log$ means $\log_{10}$, the answer would be $x=0.01.$