# Thread: Bit rusty on my log rearranging

1. ## Bit rusty on my log rearranging

Can you help me with these

Rewrite these relations in index form (that is, without using logarithims):

(a) logα (x + y) = logα x + logα y
(b) log ₁₀ x = 3 + log ₁₀ y
(c) log₃ x = 4log₃y
(d) 2log₂x + 3log₂y – 4log₂z = 0
(e) xlogα 2 = log α y
(f) log α x - log α y = n log α z
(g) ˝log₂ x + = ⅓log₂ y – 1
(h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)

2. Originally Posted by Joel
Can you help me with these

Rewrite these relations in index form (that is, without using logarithims):

(a) logα (x + y) = logα x + logα y
(b) log ₁₀ x = 3 + log ₁₀ y
(c) log₃ x = 4log₃y
(d) 2log₂x + 3log₂y – 4log₂z = 0
(e) xlogα 2 = log α y
(f) log α x - log α y = n log α z
(g) ˝log₂ x + = ⅓log₂ y – 1
(h) 2 log₃(2x + 1) = 3 log₃ (2x – 1)
$a)log_{a}(x+y) = log_{a} x + log_{a} y$

$log_{a} (x+y) = log_{a} xy$ I use $log A + log B = log (AB)$

$(x+y) = xy$

$b) log_{10} x =3 + log_{10} y$

$x=10^{3+log_{10} y }$

$x= 10^3 (10^{log_{10} y })$

$x= 1000(y)$ since $A^{log_{A} c} = c$

$e)xlog_{a} 2 = log_{a} y$

$log_{a}2^x= log_{a} y$ since $(a) log c = log c^a$

$2^x = y$

(g) ˝log₂ x + = ⅓log₂ y – 1

$g)1/2log_{2}x = 1/3log_{2} y -1$

$log_{2} x^{1/2} - log_{2} y^{1/3} = -1$

$log_{a} y^{1/3} - log_{2} x^{1/2} = 1$ multiply with -1

$log_{a} \dfrac{y^{1/3}}{x^{1/2}} = 1$

$\dfrac{y^{1/3}}{x^{1/2}} = a$

3. some log rules I attach them because in sometimes latex it is not clear if I use powers

4. Hello, Joel!

Some of your bases didn't show up . . . I'll use $b$.

Rewrite in index form (that is, without using logarithims):

$(a)\;\;\log_b(x+y) \:=\:\log_b(x) + \log_b(y)$
We have: . $\log_b(x+y) \:=\:\log_b(xy)$

Then: . $x+y \:=\:xy$

$(b)\;\;\log_b(x) \:= \:3 + \log_b(y)$
We have: . $\log_b(x) \;=\;\log_b(b^3) + \log_b(y) \;=\;\log_b(b^3y)$

Then: . $x \;=\;b^3y$

$(c)\;\;\log_b(x) \:=\: 4\log_b(y)$
We have: . $\log_b(x) \;=\;\log_b(y^4)$

Then: . $x \;=\;y^4$

$(d)\;\;2\log_b(x) + 3\log_b(y) - 4\log_b(z) \:=\:0$
We have: . $\log_b(x^2) + \log_b(y^3) - \log_b(z^4) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{x^2y^3}{z^4}\right) \:=\:0$

Then: . $\frac{x^2y^3}{x^4} \:=\:1$

$(e)\;\;x\log_b(2) \:=\: \log_b(y)$
We have: . $\log_b(2^x) \:=\:\log_b(y)$

Then: . $2^x\:=\:y$

$(f)\;\;\log_b(x) - \log_b(y) \:=\: n\log_b(z)$

We have: . $\log_b(x) - \log_b(y) - \log_b(z^n) \:=\:0 \quad\Rightarrow\quad \log_b\left(\frac{xy}{z^n}\right) \:=\:0$

Then: . $\frac{xy}{z^n} \;=\;1$

$(g)\;\;\tfrac{1}{2}\log_b(x) \:=\: \tfrac{1}{2}\log_b(y) - 1$

We have: . $\log_b\left(x^{\frac{1}{2}}\right) \;=\;\log_b\left(y^{\frac{1}{2}}\right) - \log_b(b)$

. . $\log_b\left(\sqrt{x}\right) - \log_b\left(\sqrt{y}\right) + \log_b(b) \;=\;0 \quad\Rightarrow\quad \log_b\left(\frac{\sqrt{x}}{\sqrt{y}}\cdot b\right) \;=\;0$

Then: . $\frac{b\sqrt{x}}{\sqrt{y}} \:=\:1$

$(h)\;\;2\log_b(2x + 1) \:=\: 3\log_b (2x – 1)$

We have: . $\log_b(2x+1)^2 \;=\;\log_b(2x-1)^3$

Then: . $(2x+1)^2 \;=\;(2x-1)^3$