[MOVED] exponents and logs: solving equations

[Luanne]

could someone please help me with these problems

a.) 3^(log3(3x+4)=5

b.) 5^(2log5(2x-7)=16

c.) 9^(x+2)=240+9^x

 

[soroban]

Re: exponents and logs

Hello, Luanne!

For the first two, you're expected to know that: \(\displaystyle \:b^{\log_bx}\:=\:x\)

\(\displaystyle a)\;3^{\log_3(3x+4)}\:=\:5\)

The equation becomes: \(\displaystyle \:3x\,+\,4\:=\:5\;\;\Rightarrow\;\;3x\:=\:1\;\;\Rightarrow\;\;x\,=\,\frac{1}{3}\)

With log equations, we must always check for extraneous roots.

. . And \(\displaystyle x\,=\,\frac{1}{3}\) is a valid root.

\(\displaystyle b)\;5^{2\cdot\log_5(2x-7)}\:=\:16\)

The left side is: \(\displaystyle \:5^{2\cdot\log_5(2x-7)}\:=\:5^{\log_5(2x-7)^2} \:=\2x\,-\,7)^2\)

The equation becomes: \(\displaystyle \2x\,-\,7)^2\:=\:16\;\;\Rightarrow\;\;2x\,-\,7\:=\:\pm4\;\;\Rightarrow\;\;2x\:=\:7\,\pm\,4\)

And we have two roots: \(\displaystyle \:\begin{array}{cc}2x\:=\:7\,+\,4 & \:\Rightarrow\: & x\,=\,\frac{11}{2} \\ \\ 2x\:=\:7\,-\,4 & \:\Rightarrow\: & x\,=\,\frac{3}{2}\end{array}\)

But \(\displaystyle x\,=\,\frac{3}{2}\) is extraneous . . . The only root is: \(\displaystyle \,x\,=\,\frac{7}{2}\)

\(\displaystyle c)\;9^{x+2}\:=\:240\,+\,9^x\)

This is a tricky one!

We have: \(\displaystyle \9^x)(9^2) \:=\:240\,+\,9^x\;\;\Rightarrow\;\;81\cdot9^x\:=\:240\,+\,9^x\)

. . \(\displaystyle 81\cdot9^x\,-\,9^x\:=\:240\;\;\Rightarrow\;\;80\cdot9^x\:=\:240\;\;\Rightarrow\;\;9^x\:=\:3\)

. . \(\displaystyle (3^2)^x\:=\:3\;\;\Rightarrow\;\;3^{2x}\:=\:3^1\;\;\Rightarrow\;\;2x\:=\:1\)

. . Therefore: \(\displaystyle \,x\,=\,\frac{1}{2}\)