$\displaystyle \text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)$
I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.
$\displaystyle \text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)$
I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.
I got $\frac{6(\log 2)(\log 2)+5(\log 2)(\log 3)+6(\log 3)(\log 3)}{6(\log 2)(\log 2)+5(\log 2)(\log 3)+(\log 3)(\log 3)}$
Is it right saying it $\frac{1}{5(\log 3)(\log 3)}+1$
Hello, NameIsHidden!
I'm getting incorrect answers and I don't know why.
[I know ... I sound like a kid in Algebra I, but I can't help it.]
$\displaystyle \text{If }a\,=\, \log_{12}18,\;b\,=\,\log_{24}54.\;\text{ then find the value of }ab+5(a-b)$
$\displaystyle \begin{array}{cccccccc}a \:=\: \log_{12}18 &\Rightarrow& 12^a \:=\: 18 &\Rightarrow& (2^23)^a \:=\: 18 &\Rightarrow& 2^{2a}3^a \:=\: 18 & [1] \\ \\ b \:=\: \log_{24}54 &\Rightarrow& 24^b \:=\: 54 &\Rightarrow& (2^33)^b \:=\: 54 &\Rightarrow& 2^{3b}3^b \:=\: 54 & [2] \end{array}$
Divide [2] by [1]: .$\displaystyle \frac{2^{3b}3^b}{2^{2a}3^a} \:=\:\frac{54}{18} \quad\Rightarrow\quad 2^{3b-2a}3^{b-a} \:=\:3 \:=\:2^03^1$
Equate exponents: .$\displaystyle \begin{Bmatrix}3b-2a &=& 0 \\ b-a &=& 1 \end{Bmatrix}$
Solve the system: .$\displaystyle \begin{bmatrix}a \:=\:\text{-}3 \\ b \:=\:\text{-}2 \end{bmatrix} \;\; \color{red}{??}$
I checked with the solution, the result of the expression is $1$.
While taking Soroban's explanation, I see that he missed something.
Actually $a=-3$ or $a=3$ and $b=-2$ or $b=2$.(By substituting two equations in two different ways)
The final result is either $1$ or $11$.
Are there any other ways to compute it even accurate?