Mathematics: Rewrite this logarithmic equation in exponential form

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19.06.2022

Rewrite this logarithmic equation in exponential form

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Mathematics

Rewrite the logarithm expression into exponential form

Step-by-step answer

P Answered by PhD

$\frac{1}{25}$ = $5^{-2}$

Step-by-step explanation:

Using the rule of logarithms

$log_{b}$ x = n ⇒ x = $b^{n}$ , thus

$log_{5}$ ( $\frac{1}{25}$ ) = - 2, then

$\frac{1}{25}$ = $5^{-2}$

Mathematics

Rewrite the logarithm expression into exponential form

Step-by-step answer

P Answered by PhD

$\frac{1}{25}$ = $5^{-2}$

Step-by-step explanation:

Using the rule of logarithms

$log_{b}$ x = n ⇒ x = $b^{n}$ , thus

$log_{5}$ ( $\frac{1}{25}$ ) = - 2, then

$\frac{1}{25}$ = $5^{-2}$

Mathematics

Rewrite the logarithm as an exponential equation: In(X)= 7

Step-by-step answer

P Answered by PhD

e^7=x

Step-by-step explanation:

Given the logarithm function

$In\left(x\right)=\:7$

For logarithmic equations, $\log _b\left(x\right)=y$ is equivalent to b^y=x such that

$x0\:,\:b0$ and $b\ne 1$

In current case:

$b=e\:$

$\:x=x$

$\:y=7$

Substituting the values of $b\:,\:x\:$ and into the equation b^y=x

e^7=x

Mathematics

Rewrite the logarithm as an exponential equation: In(X)= 7

Step-by-step answer

P Answered by PhD

e^7=x

Step-by-step explanation:

Given the logarithm function

$In\left(x\right)=\:7$

For logarithmic equations, $\log _b\left(x\right)=y$ is equivalent to b^y=x such that

$x0\:,\:b0$ and $b\ne 1$

In current case:

$b=e\:$

$\:x=x$

$\:y=7$

Substituting the values of $b\:,\:x\:$ and into the equation b^y=x

e^7=x

Mathematics

Question 1 In this unit, you learned about common logarithms. Since most calculators can be used to find the values of common logarithms only, it’s desirable to be able to convert logarithms with bases other than 10 to common logarithms. This conversion can be done using the change of base formula. Part A Rewrite the logarithmic equation y = logbm in exponential form. Part B Using a log with base c , take the log of both sides of your answer from part A. Then use the power property and solve for y . Part C The original equation was y=logbm . Use

Step-by-step answer

P Answered by Master

Answer and Step-by-step explanation:

Part A: Exponential equation is the "opposite" of logarithmic equation, so:

$y=log_{b}m$

$b^{y}=m$

Part B: Using log with base c:

$log_{c}b^{y}=log_{c}m$

Power property of logarithm states that if the anti-logarithm is elevated at a power, the elevated number can be pulled in front of the logarithm:

$ylog_{c}b=log_{c}m$

Solving for y:

$y=\frac{log_{c}m}{log_{c}b}$

Part C: To facilitate the use of calculators, which only have values for the base-10 log and natural log, we use change of base formula, i.e., transform

$y=log_{b}m$

into

$y=\frac{log_{c}m}{log_{c}b}$

Part D: $(log_{3}z)(log_{z}27)$

Change of base will be:

$log_{3}z=\frac{log_{10}z}{log_{10}3}$

$log_{z}27=\frac{log_{10}27}{log_{10}z}$

Solving:

$(log_{3}z)(log_{z}27)$ = $(\frac{log_{10}z}{log_{10}3})(\frac{log_{10}27}{log_{10}z} )$

$(log_{3}z)(log_{z}27)$ = $\frac{log_{10}27}{log_{10}3}$

$(log_{3}z)(log_{z}27)$ = $\frac{log_{10}3^{3}}{log_{10}3}$

$(log_{3}z)(log_{z}27)$ = $\frac{3log_{10}3}{log_{10}3}$

$(log_{3}z)(log_{z}27)$ = 3

Part E: $log_{7}300$

Using change of base:

$log_{7}300=\frac{log300}{log7}$

$log_{7}300=\frac{2.48}{0.85}$

$log_{7}300$ ≈ 3

Mathematics

Rewrite the following equation in logarithmic form. 0.25 = 2^-2 Rewrite the following equation in exponential form. log8 (512) = 3

Step-by-step answer

P Answered by PhD

log2(0.25) = -2

512 = 8^3

Step-by-step explanation:

Use of formula: loga (b) = c ⇔ b = a^c

Rewrite the following equation in logarithmic form. 0.25 = 2^-2

log2(0.25) = -2

Rewrite the following equation in exponential form. log8 (512) = 3

512 = 8^3

Mathematics

Rewrite lnx=4 as an exponential equation rewrite as a logarithmic equation: e^8=y

Step-by-step answer

P Answered by PhD

$\bf \textit{exponential form of a logarithm} \\\\ \log_a b=y \implies a^y= b\qquad\qquad a^y= b\implies \log_a b=y \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)=4\implies log_e(x)=4\implies e^4=x \\\\\\ e^8=y\implies log_e(y)=8\implies ln(y)=8$