Solve the equation algebraically using logarithms. Round to the nearest hundredth

$${7}^{3x+9}=12$$

$${7}^{3x+9}=12$$

Madilyn Quinn
2022-10-20
Answered

Solve the equation algebraically using logarithms. Round to the nearest hundredth

$${7}^{3x+9}=12$$

$${7}^{3x+9}=12$$

You can still ask an expert for help

bigfreakystargl

Answered 2022-10-21
Author has **23** answers

Any exponential function can be represented in logarithmic form, as logarithmic functions are the inverses of exponential functions. In the same way, all logarithmic functions can be represented as exponential functions.

Given equation is

$${7}^{3x+9}=12$$

Taking log on both sides,

$$\mathrm{log}({7}^{3x+9})=\mathrm{log}(12)\phantom{\rule{0ex}{0ex}}(3x+9)\mathrm{log}7=\mathrm{log}12\phantom{\rule{0ex}{0ex}}(3x+9)=\frac{\mathrm{log}12}{\mathrm{log}7}\phantom{\rule{0ex}{0ex}}3x+9=1.277\phantom{\rule{0ex}{0ex}}3x=1.227-9\phantom{\rule{0ex}{0ex}}x=-\frac{7.773}{3}=-2.591\approx -2.59$$

The required solution is $$x=-2.59$$

Given equation is

$${7}^{3x+9}=12$$

Taking log on both sides,

$$\mathrm{log}({7}^{3x+9})=\mathrm{log}(12)\phantom{\rule{0ex}{0ex}}(3x+9)\mathrm{log}7=\mathrm{log}12\phantom{\rule{0ex}{0ex}}(3x+9)=\frac{\mathrm{log}12}{\mathrm{log}7}\phantom{\rule{0ex}{0ex}}3x+9=1.277\phantom{\rule{0ex}{0ex}}3x=1.227-9\phantom{\rule{0ex}{0ex}}x=-\frac{7.773}{3}=-2.591\approx -2.59$$

The required solution is $$x=-2.59$$

asked 2022-08-21

Differentiate

${x}^{2}\mathrm{ln}(1+{x}^{2})$

I simplified it till -

$2{x}^{2}\mathrm{ln}(x)$

Then I differentiate -

$2{x}^{2}({\displaystyle \frac{1}{x}}({\displaystyle \frac{d}{dx}}x)+\mathrm{ln}x(4x)$

$\frac{2{x}^{2}}{x}}+4x\mathrm{ln}x$

Where have I gone wrong ? Because the answer I am trying to achieve is

$\frac{2{x}^{3}}{1+{x}^{2}}}+2x\mathrm{ln}(1+{x}^{2})$

${x}^{2}\mathrm{ln}(1+{x}^{2})$

I simplified it till -

$2{x}^{2}\mathrm{ln}(x)$

Then I differentiate -

$2{x}^{2}({\displaystyle \frac{1}{x}}({\displaystyle \frac{d}{dx}}x)+\mathrm{ln}x(4x)$

$\frac{2{x}^{2}}{x}}+4x\mathrm{ln}x$

Where have I gone wrong ? Because the answer I am trying to achieve is

$\frac{2{x}^{3}}{1+{x}^{2}}}+2x\mathrm{ln}(1+{x}^{2})$

asked 2022-07-20

Let $a\u03f5{\mathbb{R}}_{+}$ and $b=\mathrm{exp}(-a).$

What is the range of $\{-{b}^{2}\mathrm{log}(b)\}$? Does it range $(-\mathrm{\infty},+\mathrm{\infty})$?

How can I show $\frac{-b\mathrm{log}(b)}{1-b}\le 1$ ?

What is the range of $\{-{b}^{2}\mathrm{log}(b)\}$? Does it range $(-\mathrm{\infty},+\mathrm{\infty})$?

How can I show $\frac{-b\mathrm{log}(b)}{1-b}\le 1$ ?

asked 2022-11-18

Evaluate each expression

$a)\text{}{\mathrm{log}}_{5}\frac{1}{25}=?$

$b)\text{}{\mathrm{log}}_{2}32=?$

$a)\text{}{\mathrm{log}}_{5}\frac{1}{25}=?$

$b)\text{}{\mathrm{log}}_{2}32=?$

asked 2022-11-11

Exponential and Logarithmic Functions

Find: ${\mathrm{log}}_{2}({x}^{2}-4x-28)=2$

Find: ${\mathrm{log}}_{2}({x}^{2}-4x-28)=2$

asked 2022-10-28

A proof of why y=mx grows faster than $y=\mathrm{ln}x$ for all $m\ne 0$

How do I prove that the linear function y=mx grows faster than the logarithmic function?

How do I prove that the linear function y=mx grows faster than the logarithmic function?

asked 2022-11-07

Is there a closed-form solution for the integral

$${\int}_{0}^{\mathrm{\infty}}{\mathrm{log}}_{2}(1+ax)\cdot {e}^{-bx}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x$$

with $a,b\ge 0$?

$${\int}_{0}^{\mathrm{\infty}}{\mathrm{log}}_{2}(1+ax)\cdot {e}^{-bx}\phantom{\rule{thickmathspace}{0ex}}\mathrm{d}x$$

with $a,b\ge 0$?

asked 2022-10-29

Consider the following non linear equation where x is the independent variable, y is the dependent variabie, and A and e are constants.

$$y=-A{x}^{B}$$

Which of the following is the resulting gradient variable in linearised form?

Select one:

$$a)log\text{}10(B)\phantom{\rule{0ex}{0ex}}b)A\phantom{\rule{0ex}{0ex}}c)exp(B)\phantom{\rule{0ex}{0ex}}d)B\phantom{\rule{0ex}{0ex}}e)log(-A)$$

$$y=-A{x}^{B}$$

Which of the following is the resulting gradient variable in linearised form?

Select one:

$$a)log\text{}10(B)\phantom{\rule{0ex}{0ex}}b)A\phantom{\rule{0ex}{0ex}}c)exp(B)\phantom{\rule{0ex}{0ex}}d)B\phantom{\rule{0ex}{0ex}}e)log(-A)$$