Question

Hi, need help with solving this logarithm.​

Answer 1

Answer:

$\log 8 - \log x + 7\log\sqrt x =\log (8x^{\frac{5}{2}})$

Step-by-step explanation:

Given

$\log 8 - \log x + 7\log\sqrt x$

Required

Express as a single expression

We have:

$\log 8 - \log x + 7\log\sqrt x$

Write 7 as an exponent

$\log 8 - \log x + 7\log\sqrt x =\log 8 - \log x + \log(\sqrt x)^7$

Rewrite as:

$\log 8 - \log x + 7\log\sqrt x =\log 8 - \log x + \log(x^{\frac{1}{2}})^7$

$\log 8 - \log x + 7\log\sqrt x =\log 8 - \log x + \log x^\frac{7}{2}$

Apply quotient and product rule of logarithm

$\log 8 - \log x + 7\log\sqrt x =\log (\frac{8*x^\frac{7}{2}}{x} )$

Apply law of indices

$\log 8 - \log x + 7\log\sqrt x =\log (8*x^{\frac{7}{2} - 1})$

Solve exponent

$\log 8 - \log x + 7\log\sqrt x =\log (8*x^{\frac{7-2}{2}})$

$\log 8 - \log x + 7\log\sqrt x =\log (8*x^{\frac{5}{2}})$

$\log 8 - \log x + 7\log\sqrt x =\log (8x^{\frac{5}{2}})$