Use properties of logarithms to find exactly value.

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}$

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}$

John Landry
2022-07-27
Answered

Use properties of logarithms to find exactly value.

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}$

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}$

You can still ask an expert for help

escobamesmo

Answered 2022-07-28
Author has **18** answers

Properties to use here:

${A}^{{\mathrm{log}}_{A}(x)}=x$

$\mathrm{log}(a)+\mathrm{log}(b)=\mathrm{log}(ab)$

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}={5}^{{\mathrm{log}}_{5}(6\ast 7)}={5}^{{\mathrm{log}}_{5}(42)}=42$

${A}^{{\mathrm{log}}_{A}(x)}=x$

$\mathrm{log}(a)+\mathrm{log}(b)=\mathrm{log}(ab)$

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}={5}^{{\mathrm{log}}_{5}(6\ast 7)}={5}^{{\mathrm{log}}_{5}(42)}=42$

Karsyn Beltran

Answered 2022-07-29
Author has **5** answers

${5}^{{\mathrm{log}}_{5}6+{\mathrm{log}}_{5}7}$

${5}^{{\mathrm{log}}_{5}42}[{\mathrm{log}}_{a}+{\mathrm{log}}_{b}={\mathrm{log}}_{ab}]$

$42[{a}^{{\mathrm{log}}_{a}x}=x]$

42 is the answer.

${5}^{{\mathrm{log}}_{5}42}[{\mathrm{log}}_{a}+{\mathrm{log}}_{b}={\mathrm{log}}_{ab}]$

$42[{a}^{{\mathrm{log}}_{a}x}=x]$

42 is the answer.

asked 2022-01-22

How can I solve $8{n}^{2}=64n{\mathrm{log}}_{2}\left(n\right)$

asked 2022-07-19

problem solving logarithmic equation and reaching an equivalence

ok so i've had a problem trying to simplify the $\mathrm{ln}[\sqrt{1+\frac{{u}^{2}}{{a}^{2}}}+\frac{u}{a}]$ and this is supposed to be equal to : $\mathrm{ln}[\sqrt{{a}^{2}+{u}^{2}}+u]$

how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?

ok so i've had a problem trying to simplify the $\mathrm{ln}[\sqrt{1+\frac{{u}^{2}}{{a}^{2}}}+\frac{u}{a}]$ and this is supposed to be equal to : $\mathrm{ln}[\sqrt{{a}^{2}+{u}^{2}}+u]$

how is this posible ?? i've tried to solve this for more than 2 hours and couldn't get to this equivalence. any suggestions ?

asked 2021-10-20

Use properties of logarithms to rewrite the expression as a sum, difference, or multiple of logarithms.

$\mathrm{ln}\frac{3x}{\sqrt{{x}^{2}-5}}$

asked 2022-07-05

Rewrite a formula in terms of exponential to the power of logarithm

I would like to rewrite the following formula, f(x).

how can I rewrite the f(x)

$f(x)=\frac{(1+2B)}{2A\sqrt{{B}^{2}+(1+2B)\frac{x}{A}}}$

as

${\mathrm{exp}}^{\mathrm{log}(f(x)}$

I would like to rewrite the following formula, f(x).

how can I rewrite the f(x)

$f(x)=\frac{(1+2B)}{2A\sqrt{{B}^{2}+(1+2B)\frac{x}{A}}}$

as

${\mathrm{exp}}^{\mathrm{log}(f(x)}$

asked 2022-04-01

How to solve

$\frac{\left(3{\mathrm{log}}_{y}5\right)\left(2{\mathrm{log}}_{y}5\right)}{\left(6{\mathrm{log}}_{y}5\right)}$ ?

asked 2022-01-07

Given $\frac{\mathrm{log}x}{b-c}=\frac{\mathrm{log}y}{c-a}=\frac{\mathrm{log}z}{a-b}$ show that ${x}^{b+c-a}\cdot {y}^{c+a-b}\cdot {z}^{a+b-c}=1$

asked 2022-04-05

Doubt on Logarithms multiplication

today I'm in doubt on calculating the follow expression${\mathrm{log}}_{4}3\cdot {\mathrm{log}}_{9}32$

Changing all to base 4: Working on:$\mathrm{log}}_{4}3\cdot \frac{{\mathrm{log}}_{4}32}{{\mathrm{log}}_{4}9$

Ending with:$\mathrm{log}}_{4}3\cdot \frac{2+{\mathrm{log}}_{4}2}{2\cdot {\mathrm{log}}_{4}3$

There's a way to simplify it more ? Also, do you know any resource explaining more on rules on every kind of operation with logs ?

Thanks in advance

today I'm in doubt on calculating the follow expression

Changing all to base 4: Working on:

Ending with:

There's a way to simplify it more ? Also, do you know any resource explaining more on rules on every kind of operation with logs ?

Thanks in advance